w.p.energy final colour

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WORK, POWER & ENERGY www.locuseducation.org

01. Concept of Energy 02 - 03

02. Work: Definition and Calculation 03 - 12

03. Work Energy Theorem and its Application 13 - 24

04. Power 25 - 31

05. Potential Energy and Energy Conservation 32 - 47

06. More on Conservative forces 48 - 54

07. A Note on Mass and Energy (Optimal) 55 - 59

08. Miscellaneous Examples 60 - 87

CONCEPT NOCONCEPT NOCONCEPT NOCONCEPT NOCONCEPT NOTESTESTESTESTES

Work, Power& Energy

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Introduction:

Any body (or an assembly of bodies) represents, in fact, a system of mass points, or particles. If a systemchanges with time, it is said that its state varies. The state of a system is defined by specifying the positions andvelocities of all constituent particles.

Experience shows that if the laws of forces acting on a system and the state of the system at a certain initialmoment are known, the motion equations can help predict the subsequent behaviour of the system.

However, an analysis of a system’s behaviour by the use of motion equations requires so much effort (due to thecomplexity of the system itself), that a comprehensive solution seems to be practically impossible. Moreover,such an approach is absolutely out of the question if the laws of acting forces are not known. Besides, there aresome problems in which the accurate consideration of motion of individual particles is meaningless (example:gas).

Under these circumstances the following question naturally comes up: are there any general principles followingfrom Newton’s laws that would help avoid these difficulties by opening up some new approaches to the solutionof the problem.

It appears that such principles exist. They are called conservation laws.

As we have already discussed, the state of a system varies in the course of time as that system moves. However,there are some quantities, state functions, which possess the very important and remarkable property of retainingtheir values with time provided you don’t change your frame. Among these constant quantities, energy, linearmomentum and angular momentum play the most significant role.

The laws of conservation of energy (There used to be a separate conservation law or more until Einstein unifiedit with energy), momentum and angular momentum fall into the category of the most fundamental principles ofphysics, (when you will study the law of conservation of linear momentum you will find that if is just a restatementof Newtons’ second law: ( )/F d mv dt= ! applied to a system. Angular momentum too, is some thing similar,conservation of energy on the other hand, is a novel concept). These laws have become even more significantsince it was discovered that they go beyond the scope of mechanics and represent universal laws of nature. Inany case, no phenomena have been observed so far which do not obey these laws. They work reliably in allquarters: in the field of elementary particles, in outer space, in atomic physics and in solid state physics.

Having made possible a new approach to treating various mechanical phenomena, the conservation laws turnedinto a powerful and efficient tool of research used by physicists. The importance of the conservation principles isdue to several reasons:

1. The conservation laws do not depend on either the paths of particles or the nature of acting forces.Consequently, they allow us to draw some general and essential conclusions about the properties ofvarious mechanical processes without resorting to their detailed analysis by means of motion equations.

2. Since the conservation laws do not depend on the nature of the acting forces, they may be appliedeven when the forces are not known. In these cases the conservation laws are the only tools forresearch remaining. This is the present trend in the physics of elementary particles.

WORK POWER & ENERGY

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3. Even when the forces are known precisely, the conservation laws can help substantially to solve manyproblems of motion of particles. Although all these problems can be solved with the use of motionequations (and the conservation laws provide no additional information in this case), the utilization ofthe conservation laws very often allows the solution to be obtained in the most straight-forward andelegant fashion, obviating cumbersome and tedious calculations.

We shall begin examining the conservation laws with the energy conservation law, having introduced the conceptof energy and work.

Section - 1 CONCEPT OF ENERGY

Energy : It is possible to give a numerical rating, called energy, to the state of a physical system. The totalenergy is found by adding up contributions from characteristics of the system such as motion of objects in it, heatcontent of the objects (though that can also be attributed to mechanical vibrations of particles), and the relativepositions of objects that interact via forces. The total energy of a closed system always remains constant. Energycan not be created or destroyed, but only transferred from one system to another.Energy comes in a variety of forms, and physicists didn’t discover all of them right away. They had to startsomewhere, so they picked one form of energy to use as a standard for creating a numerical energy scale. Onepractical approach is to define an energy unit based on heating of water. The SI unit of energy is the joule, J,named after the British physicist James Joule. One joule is the amount of energy required in order to heat 0.24 gof water by 1°C.

Note that heat, which is a form of energy, is completely different from temperature. In standard, formal terminology,there is another, finer distinction. The word heat is used only to indicate an amount of energy that is transferred,whereas thermal energy indicates an amount of energy contained in an object.

Once a numerical scale of energy bas been established for some form of energy such as heat, it can easily beextended to other types of energy. For instance, the energy stored in one gallon of gasoline can be determined byputting some gasoline and some water in an insulated chamber, igniting the gas, and measuring the rise in thewater’s temperature. (The fact that the apparatus is known as a bomb calorimeter will give you some idea ofhow dangerous these experiments are if you don’t take the right safety precautions). Here are some examples ofother types of energy that can be measured using the same units of joules.

Type of energy:• chemical energy released by burning • energy required to break an object• energy required to melt a solid substance • chemical energy released by digesting

food• raising a mass against the force of gravity • nuclear energy released in fission, etc.

Textbooks often give the impression that a sophisticated physics concept was created by one person who had aninspiration one day, but in reality it is more in the nature of science to rough out an idea and then gradually refineit over many years. The idea of energy was tinkered with from the early 1800’s onwards, and new types ofenergy kept getting added to the list.

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To establish a new form of energy, a physicist has to1. show that it could be converted to and from other forms of energy, and2. show that it is related to some definite measurable property of the object, for example its temperature,

motion, position relative to another object, or being in a solid or liquid state.

For example, energy is released when a piece of iron is stored in water, so apparently there is some form of energyalready stored in the iron. The release of this energy can also be related to a definite measurable property of thechunk of metal: it turns reddish-orange. There has been a chemical change in its physical state, which we callrusting.Although the list of types of energy kept getting longer and longer, it was clear that many of the types were justvariations on a theme. There is an obvious similarity between the energy needed to melt ice and to melt butter, orbetween rusting of iron and many other chemical reactions. All the types of energy can be reduced to a very smallnumber by simplifications.

Section - 2 WORK: DEFINITION AND CALCULATION

The concept of work:The mass contained in closed system is a conserved quantity, but if the system is not closed, we also have ways ofmeasuring the amount of mass that goes in or out.

We often have a system that is not closed, and would like to know how much energy comes in or out. Energy,however, is not a physical substance like water, so energy transfer can not be measured by the same kind of meterwhich we used to measure flow of water. How can we tell, for instance, how much useful energy a tractor canput out on one tank of gas?

The law of conservation of energy guarantees that all the chemical energy in the gasoline will reappear in someform, but not necessarily in a form that is useful for doing farm work. Tractors, like cars, are extremely inefficient,and typically 90% of the energy they consume is converted directly into heat, which is carried away by the exhaustand the air flowing over the radiator. We wish to distinguish energy that comes out directly as heat from the energythat serves to accelerate a trailer or to plow a filed, so we defined a technical meaning of the ordinary work toexpress the distinction.

Definition of work:Work is the amount of energy transferred into or out of a system, not taking into account energytransferred by heat conduction.

[Based on this definition, is work a vector, or a scalar? What are its units?]

The conduction of heat is to be distinguished from heating by friction. When a hot potato heats up your hands byconduction, the energy transfer occurs without any force, but when friction heats your car’s brake shoes, there isa force involved. The transfer of energy with and without a force are measured by completely different methods,so we wish to include heat transfer by frictional heating under the definition of work, but not heat transfer byconduction. The definition of work could thus be restated as the amount of energy transferred by forces.

Work done by a constant force:Work done by a constant force is defined as product of the force and the component of the displacementalong the direction of the force.

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4F

S mm

fig 5.1.

F

Consider the situation shown in figure 5.1. A constant force F is applied on a block of mass m along the horizontaldirection. If the block moves by a distance S on the horizontal surface on which it is placed, as shown in figure,then the work done by the force F is defined as

⋅=w F S ...(1).

If the force and the displacement are not along the same direction, as shown in figure 5.2, then work done by forceF is calculated by multiplying the force and the component of the displacement along the force, as shown in figure5.3, therefore, for the given case, work done by force F is

F

S mm

fig. 5.2

F

θ

S cos θ

mm SθF(const.)

fig. 5.3

cosw F S θ= ⋅

⇒ ⋅ ⋅= cosw F S θ ...(2)

Here you should note that work done by the force F can also be written as

( cos )w F Sθ= ⋅

and we know that Fcosθθθθθ is the component of F along the displacement, as shown in figure 5.4.

F cos θmm

Sθ

F(const.)

work done by = force along displacement × displacement

F

fig. 5.4

Hence, work done by a constant force can also be defined as the product of the displacement and thecomponent of the force along the displacement.

In vector form equations (1) and (2) can be generalized as

⋅=!!

w F S ...(3)

Hence, work done by a constant force is the scalar (dot) product of the force and the displacement. Here

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I would like to emphasize that !S is the displacement of the point of application of the force

!F .

Note: For a constant force !

F , in equation (3) you should notice the following:

• When i.e., 90 , 0.fF S, wθ⊥ = ° =!!

• When (0,90 ), 0.fwθ ∈ ° >

• When (90 ,180 ), 0.fwθ ∈ ° ° <

• When 180 , fw F Sθ = ° = − ⋅

• When 0 , fw F Sθ = ° = ⋅

• In a closed path work done by a constant force is zero. ( 0).S =! !

∵( fw denotes work done by the constant force F and S is the magnitude of the displacement S

!)

Equation (3) refers only to the work done on the particle by a particular force F!

. The work done on the particleby the other forces must be calculated separately. The total work done on the particle is the sum of the work doneby the separate forces.

( = 0°)θF SF

Sm

fig. 5.5

When θ is zero, as shown in figure 5.5, the work done by F!

issimply F × S, in agreement with equation (1). Thus, when aconstant horizontal force draws a body horizontally, or when aconstant vertical force lifts a body vertically, the work done bythe force is the product of the magnitude of the force by thedistance moved.

When θ is 90°, as shown in figure 5.6, the force has nocomponent in the direction of motion. That force then does nowork on the body. For instance, the vertical force holding abody a fixed distance off the ground does no work on the body,even if the body is moved horizontally over the ground. Also thecentripetal force acting on a body in motion does no work onthat body because the force is always at right angles to the direction in which the body is moving.

F

Sm

fig. 5.6

( = 90°)θF S

Of course, a force does no work on a body that does not move, for its displacement is then zero.

As I have already mentioned, the work done by a constant force can be calculated in two different ways: Eitherwe multiply the magnitude of the displacement by the component of the force in the direction of the displacementor we multiply the magnitude of the force by the component of the displacement in the direction of the force.These two methods always give the same result.

Work is a scalar, although the two quantities involved in its definition, force and displacement, are vectors. Wedefine the scalar product of two vectors as the scalar quantity that we find when we multiply the magnitude of onevector by the component of a second vector along the direction of the first. Equation (3) shows that work is sucha quantity.

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Work can be either positive or negative. If the particle on whicha force acts has a component of motion opposite to the directionof the force, the work done by that force (Fig. 5.7) is negative.This corresponds to an obtuse angle, between the force anddisplacement vectors. For example, when a person lowers anobject to the floor, the work done on the object by the upwardforce of his hand holding the object is negative.

F

Sm

fig. 5.7

( > 90°)θθ

Our special definition of the word work does not correspond to the daily usage of the term. This may be confusing.A person holding a heavy weight at rest in the air may say that he is doing hard work- and he may work hard in thephysiological sense-but from the point of view of physics we say that he is not doing any work. We say thisbecause the applied force causes no displacement. The word work is used only in the strict sense of equation (3).

The unit of work is the work done by a unit force in moving a body a unit distance in direction of theforce. In the mks system the unit of work is 1 newton-meter, called 1 joule.

Work done by gravity (near earth surface):Near the earth’s surface the gravitational force acting on a body due to attraction of the earth can be assumed tobe constant. Therefore work done by gravity can be calculated using equation (3), whatever be the path of motionof the body.

S cosθ

path

2

1

h

fig. 5.8

mg!

mg!mg!mg!

S!

S

S cosθ

mgpath

2

1

h

fig. 5.9

θ

Consider the situation shown in figure 5.8. A particle of mass m is moved on an arbitrary path in a vertical plane.As the particle is moved from point 1 to point 2, its weight acting on it at different positions during its motion is alsoshown in figure. From figure it is clear that work done by gravity can be found by multiplying mg (magnitude of theforce) by the downward displacement h of the body because the component of the displacement of the bodyalong the force is h only. If you are not satisfied with the discussion above and want to calculate it mathematically,then you can proceed according to the following way:

Work done by gravity when particle moves from point 1to point 2 as shown in figure 5.9 ismg s= ⋅! !gw

cosmg s θ= ⋅ [ is constant]mg!∵

.mg h=

= weight of the particle × downward displacement of the particle

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Hence, we have got the same result as we had predicted earlier.

Therefore, when a particle comes down by a distance h through any path, work done by gravity is given as

(downward motion)

=gw mgh...(4)

You should note that this work done is independent of the path of the particle. You should also note that “mgh” iswork done by gravity only, not by all forces acting on the particle when it moved from point 1 to point 2.

What would be the work done by gravity when the particle moves from point 2 to point 1? In this case thedisplacement of the particle has a component of length h in the vertical direction but this component is in theopposite direction of ,mg! hence, work done by gravity for this case is

(upward motion)

= −gw mgh...(5)

You should note that this expression is also independent of the path followed by the particle while moving frompoint 2 to point 1.

Now, let me propose a common equation for the equations (4) and (5). When a particle of mass m moves nearearth surface, work done on it by gravity is given by

= − ∆gw mg h ...(6)

where ∆h is the change in height of the particle. If the particle goes up by a height h then ∆h = +h andgw mgh= − , which is in accordance with equation (5). If the particle comes down by a height h then ∆h = –h and

work done by gravity, ,gw mgh= + which is in accordance with the equation (4). Hence, for all cases we can useequation (6). The only restriction while using equation (4), (5) or (6) is that ‘mg’ must be uniform over the path forwhich work done is required.Note that if the particle moves from 1 to 2 and then from 2 to 1, then work done by gravity is zero. Hence, we cansay that work done by gravity in a closed path is zero.

Work done by a variable force:

�θ

2

1

d s!

fig 5.10.

When the force varies over the path of motion of a particle (itcan vary in magnitude or direction or both), then to calculatework done by the force we divide the path of the particle inmany infinitesimally small intervals and calculate the work donein each interval and by adding the work done for all these smallintervals we get the work done over the segment of the pathunder consideration. In figure 5.10, such an interval, ,d s! isshown. As the interval is very small, we can assume that in thisinterval force is constant and hence work done by F

! in this

interval can be written as

= ⋅! !dw F ds ...(7)

where ds! is the displacement of the particle for this interval. As the particle moves from point 1 to point 2 on thepath shown above, the net work done by the force F

! is calculated by adding work done in all subintervals of the

path from point 1 to point 2. Therefore, net work done by F!

is

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sum of all 'sw dw=

⇒ w dw= ∫

⇒ = ⋅∫2

1

! !w F d s ...(8)

You should note that when you apply equation (8) for a constant force you get the same result as given by equation(3).

For a one dimensional case equation (8) can be modified by replacing d s! with dx and replacing F!

with F(x).Therefore, we get,

( )f

i

x

xw F x dx= ⋅∫ ...(9)

For a 3-dimensional case equation (8) can be expanded as follows:

fin

in

w F ds= ⋅∫! !

( ) ( )ˆ ˆˆ ˆ ˆ ˆ.fin

x y zin

F i F j F k dxi dyj dzk= + + + +∫

( )fin

x y zin

F dx F dy F dz= ⋅ + ⋅ + ⋅∫

⇒

f f f

i i i

x y z

x y zx y z

w F dx F dy F dz= ⋅ + ⋅ + ⋅∫ ∫ ∫ ...(10)

If components of F!

along x, y and z are constant, then equation (10) reduces to

∆ ∆ ∆x y zw F x F y F z= ⋅ + ⋅ + ⋅ ...(11)

There is an alternate way of looking at work done by variable forces which would prove to be a powerful methodas we will proceed with the topic. Suppose we consider F

! as the sum of its components along the tangent and the

normal to the path of motion of the particle. In that case equation (8) gives

fin

in

w F d s= ⋅∫! !

( )ˆ ˆ and are unit vectors along the tangent

and the normal to the path, respactively,

at the point under consideration

ˆ ˆfin

t nin

t n

F t F n ds = ⋅ + ⋅ ⋅

∫!

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ˆ ˆ( ) ( )fin

t nin

F t d s F n ds = ⋅ + ⋅ ∫! !

⇒ ⋅∫fin

tin

w = F ds ˆandˆ d s t d s dsn ⋅ = ⊥ ! !#∵ ...(12)

Hence, work is done only by the tangential component of a force.

Example – 1 WORK DONE BY SPRING FORCE

A light spring of spring constant k is stretched from an initial deformation ix to a final deformation fx quasistatically(i.e., equilibrium is always maintained). Find the work done by

(a) spring force (b) external agent

Solution: In figure 5.11 the spring is shown at some arbitrarydeformation x. As the spring always exerts restoringforce, at the shown moment it is applying a force inthe opposite direction of x. Therefore,

= −spF kx (always true) where k is the spring

natural length

xi

xf

xdx

FF

fig 5.11.

ext

sp

constant of the spring.

To increase the deformation the external must exert a force in the direction of x and the force exertedby the external agent, ,extF must have a magnitude equal to that of spF because equilibrium is alwaysmaintained. Therefore,

= +extF kx

true only if equilibrium is maintained

If the spring is further deformed by dx from the shown position then work done by the spring force is

( )spdw kx dx= − ⋅

and work done by external agent is

( )extdw kx dx= + ⋅

While deforming the spring from ix to ,fx we get, total work done by the spring,

f

i

x

sp spx

w dw k x dx= = − ⋅∫ ∫

⇒2 21 1

2 2sp f iw kx kx = − − ...(13)

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and total work done by external agent

f

i

x

ext extx

w dw k x dx= = + ⋅∫ ∫

⇒ 2 21 12 2ext f iw kx kx= − ...(14)

Initially if the spring has its natural length then substituting 0,ix = equation (13) and (14) give

= − 212sp fw kx ...(15)

and = + 212ext fw kx ...(16)

Note: You must notice the following

• Equations (13) and (15) are true for all cases but equations (14) and (16) are true only if equilibriumis maintained.

• Work done by the spring is independent of path, it depends only upon initial and final position.[See equation (13) and (15)]

• You should note that ( )F x dx⋅∫ denotes area under F(x)curve when it is plotted against x.

• From figure. 5.12,

shaded areaspw =

fig 5.12.

xi xf

A1

A2 W� kxi

� kxf

F x( )

x

F x kxF

( )= � = sp

sp

1 2A A= +

( )( )i f ikx x x= − − +1 ( )( )2 f i f ikx kx x x− + −

2 21 1

2 2f ikx kx = − −

• If we put i fx x= in equation (13), then we get 0.spw = Therefore, we can say that work done byspring force in a closed path is zero.

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Work Done by Two or More than two forces:

Suppose there are n forces 1 2, ,........., nF F F! ! !

acting on a particle of mass m, as shown in figure 5.13. If the particlesuffers a displacement ds! in the next time interval dt, then the net work done on the particle is the sum of workdone by all the forces acting on it. Hence, we have net work done on the particle

1 2 ....... ndw dw dw dw dw= + + + +

1 2 3 ....... nF ds F ds F ds F ds= ⋅ + ⋅ + ⋅ + + ⋅! ! ! !! ! ! !

F1

F3 F2

Fn

m

fig. 5.13

! !

!

!

where i idw F ds= ⋅! ! is work done by the force iF

!.

For a finite displacement, net work done on the particle is

......= + + +1 2 nw w w w ...(17)

1 2 .......... nF ds F ds F ds= ⋅ + ⋅ + + ⋅∫ ∫ ∫! ! !! ! !

( )1 2 ...... nF F F ds= + + + ⋅∫! ! ! !

⇒! !

netw F ds= ⋅ ...(18)

Therefore, if many forces are acting on a particle, then we have two ways to find out net work done on theparticle:

1. find work done by each force individually and then find the sum of work done by all forces;

2. find the net force, sum of all forces, acting on the particle and then find work done by this net force.

When you are applying the above approach for a particle then no attention is needed but while applying thisconcept on a system of particles or on a body you need to be cautious. If points of application of different forceshave different displacements, then equation (17) can not be reduced to equation (18). Hence, in such a case youare left with only one option and that is of using equation (17).

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TRY YOURSELF - I

Q. 1 No work is done by a force on an object if(a) the force is always perpendicular to its velocity(b) the force is always perpendicular to its acceleration(c) the object is stationary but the point of application of the force moves on the object(d) the object moves in such a way that the point of application of the force remains fixed.

Q. 2 Find the work a boy of weight 55 kg has to do against gravity when climbing from the bottom to thetop of a 3.0 m high tree. (g = 10 m/s²).

Q. 3 A body is thrown on a rough surface such that friction force acting on it varies linearly with distancetraveled as f = ax + b. Find the work done by the friction on the box if before coming to rest the boxtravels a distance s.

Q. 4 A force is given by 2F kx= , where x is in meters and 210 /k Nt m= . What is the work done by this forcewhen it acts from x = 0 to x = 0.1 m?

Q. 5 A body is acted upon by a force which is inversely proportional to the distance covered. The work donewill be proportional to :

(a) s (b) s²(c) s (d) ln s

Q. 6 A force F acting on a particle varies with the position xas shown in figure. Find the work done by this force indisplacing the particle from(a) 2m to 0x x= − =

10

x m( )

F n( )

�10

�22

(b) 0 to 2m.x x= =

Q. 7 Two unequal masses of 1 kg and 2 kg are attached to the two ends of a light inextensible string passingover a smooth pulley as shown in figure. If the system is released from rest, find the work done by thestring on both the blocks in 1 s. (Take g = 10 m/s²)

1 kg

2 kg

Q. 8 A particle moves from a point 1ˆ ˆ(2 ) (3 )r m i m j= +! to another point 2

ˆ ˆ(3 ) (2 )r m i m j= +! during which a

certain force ˆ ˆ(5 ) (5 )F N i N j= +$!

acts on it. Find the work done by the force on the particle during thedisplacement.

Q. 9 A block of mass m moving at a speed v compresses a spring through a distance x before its speed ishalved. Find the spring constant of the spring.

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Section - 3 WORK ENERGY THEOREM AND ITS APPLICATION

As we have already discussed that the normal component of a force does no work, the work is done only by thetangential component of a force. The tangential component is related to rate of change of speed of the particle. By

the Newton’s 2nd law, ,tdvF mdt

=∑ where v is speed of the particle and tF∑ denotes the sum of tangentialcomponents of all forces acting on the particle. We can now think of speed as a function of the distance smeasured along the curve (as shown in figure 5.14) and apply the chain rule for derivatives:

dv dv ds dvvdt ds dt ds

= ⋅ = ⋅

dsdtv=

s

startingpoint fig. 5.14

∴ tdvF m vds

∑ = ⋅ ⋅

where we have used the fact that dsdt is just the speed v. The work done by resultant force is thus

fin fin

net tin in

dvw F ds mv dsds

= ⋅ = ⋅ ⋅∑∫ ∫

fin fin

in in

mv dv m v dv= ⋅ = ⋅∫ ∫

or 2 21 12 2net f iw mv mv= − ...(19)

The quantity 212

mv has the same unit as that of work and hence it is a form of energy. As it depends upon the

speed of the particle, it is defined as the kinetic energy, k, of the particle. Hence equation (19) can be rewrittenas

net f iw k k= −

⇒ net = ∆w k ...(20)

Therefore, net work done on a particle is equal to the change in kinetic energy of the particle. Thistheorem is known as the work energy theorem.

If positive work is done on a particle then kinetic energy of the particle increases. If work done on a particle isnegative then kinetic energy of the particle decreases. If no work is done on a particle then kinetic energy of theparticle remains unchanged and hence speed of the particle remains constant. At this juncture you should recall the

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very basic definition of work. We had defined it as energy transferred by forces and here the work energytheorem proves this definition. Now, it is clear that all the work done on a body goes in increasing the kineticenergy of the body.Using equation (12) and (20), we can get

dw dk=

tF ds= ⋅

⇒ =tdkFds ...(21)

Therefore, the derivative of kinetic energy with respect to distance covered along the path gives the tangentialcomponent of the net force acting on a particle. The normal component of the net force can be found by multiplyingthe mass of the particle by the centripetal acceleration of the particle. We have already learnt that, in order tochange the speed, we need a tangential component of the force (in circular motion). It is in accordance withequation (21). If the tangential force is zero, then no work is done on a particle and hence its kinetic energyremains unchanged or we can say that the speed of the particle remains unchanged. Therefore to change thespeed there must be a component of the net force along the tangent to the path and this is exactly what wediscussed while covering non uniform circular motion.

Example – 2

A small ball released from rest from a height hon a smooth surface of varying inclination, asshown in figure 5.15. Find the speed of the ballwhen it reaches the horizontal part of the surface.

smooth

h

fig. 5.15

Solution: Let the speed of the ball when itreaches the horizontal part of thesurface be 0v , as shown infigure 5.16(b).

fig. 5.16(a)

v

N

mg

m

v0

fig. 5.16(b)

The position of the ball at some arbitrary point of the inclined part of the surface is shown infigure 5.16(a). During the entire journey there are only two forces acting on the ball:(1) Its weight; (2) Normal contact force from the surface. Applying the work energy theorem betweenthe moments when it was released from the rest and when it reaches the horizontal surface, we get.

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netw k= ∆

⇒0Ngw w+

0ifk k= −work done by gravity;

work done by normal contact forceg

N

ww

→ →

⇒ 20

10 02

mgh mv+ = −from equation (6),

gw mgh =

⇒ = 20v gh ...(22)

As the normal contact force is always perpendicular to the direction of motion, work done by it iszero.When a ball falls freely from rest by a distance h, its speed is the same as obtained in equation (22).During free fall the only force doing work is gravity.

Example – 3

In the previous problem suppose that an uncompressedspring is fixed on the horizontal part of the surface, asshown in figure 5.17. Now, if the particle is releasedfrom rest from the position shown in the figure, find themaximum compression in the spring if its spring constantis k.

fig. 5.17

m

k

h

Solution: When compression in the spring is maximum, speed of the ball must be zero at that moment, as shownin figure 5.18 (c). If 0x be the maximum compression in the spring then applying work energy theoremon the ball between the instants when it was released from rest and when the compression in thespring is maximum, we get

v

N

mg

m N

mg

v0

x 0

v=0fig. 5.18(a) fig. 5.18(b) fig. 5.18(c)

netw k= ∆

⇒0Ngw w+

0fspw k+ =

0ik− is work done by springspw

⇒ 20

1 02

mgh kx+ − =

⇒ 02mghx

k=

You should try to prove that the ball would bounce back to the same point from which it was released.

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Example – 4

In the previous example suppose the spring is removed from the horizontal part of the surface and this part is maderough. If the coefficient friction between the rough part of the surface and the ball is µ, find the distance coveredby the ball on the horizontal part of the surface before it comes to rest.

Solution: Let the ball comes to rest having covered a distances on the horizontal part of the surface. During themotion of the ball only three forces act on the ball:(1) Gravity; (2) Normal contact force; (3) Frictionalforce.

v

N

mg (a)Fig. 5.19Applying work energy theorem between the instantswhen the ball was released from rest and the ballcomes to rest on the horizontal part, we get,

netw k= ∆

⇒ g N fr f iw w w k k+ + = −

(b)

v

N

mg

µmgsmoothrough

Fig. 5.19

⇒ 0 0 0mgh mg sµ+ + − ⋅ = −

⇒ = hsµ

Example – 5A block of mass m suspended vertically by a light spring of springconstant k is released from rest when the spring is in its naturallength state, as shown in figure 5.20. Find the maximum elongationin the spring.

m

k naturallength

�released from rest�Fig 5.20.

Solution: As the block is released from rest it falls downwardsdue to its own weight but as it moves downwards itsdownward acceleration decreases due to the upwardforce exerted by the spring. Initially the spring forceis zero, hence the downward acceleration is maximum(=g). But as the spring elongates the net downward force decreases and hence the downwardacceleration also decreases. After some time the upward spring force becomes equal to the weight ofthe block. This position is known as equilibrium position because net force in this position is zero. Youshould note that at this position speed of the block is not zero. In fact, at this position speed of theblock is maximum, because before arriving to this position spring force is smaller than the weight ofthe block and hence the speed of the block is increasing in the downward direction.

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When the block reaches the equilibrium position, the net force on it becomes zero and hence itsacceleration also becomes zero. Therefore, speed becomes maximum because it can not increasefurther. Although acceleration of the block is zero in this position, it will continue moving in the downwarddirection due to its downward velocity.

mga = g

v=0

x0mgk

kx x

a v(increasing speed)

kx0

mg vmax

kxa

x

mg

v (decreasing speed)

2kx0

mg

=2mg

maximumelongationposition

equilibriumposition

natural lengthposition

fig 5.21.

a = 0

a = g

2x =0

2mgk

=xmax

v=0

As the block passes the equilibrium position, the spring force exceeds the weight of the block andhence the block starts decelerating and then after moving some distance with decreasing speed, itmomentarily comes to rest. Obviously when the ball comes to rest, elongation in the spring is maximum.If maxx is the maximum elongation in the spring then applying work energy theorem on the blockbetween the moment when it was released and when the elongation is maximum, we get

netw k= ∆

⇒ g sp f iw w k k+ = −

⇒ 2max max

1 0 0 02

mg x kx+ ⋅ − = − =

⇒ max2mgx

k=

⇒ max 02x x=

where 0x is the elongation in the spring in the equilibrium position.

Note: � At maxx x= block momentarily comes to rest.

• At this position velocity of the block is zero but acceleration is not zero. It has an upward accelerationof magnitude g, as shown in figure 5.21.

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• When elongation is maximum in the spring its speed is zero which can be proved very easily if youproceed mathematically rather than analyzing exactly what’s happening there. When elongation in thespring is maximum, distance of block from natural length state, x, as shown in figure 5.21, is alsomaximum. Hence, we have

0 0.dx vdt

= ⇒ =

Example – 6

A small ball of mass m is suspended by means of a light thread oflength l. When the ball is hanging vertically it is given a horizontalspeed u, as shown in figure 5.22. Find the speed of the ball andtension in the thread supporting the ball when thethread makes an angle θ with the vertical.

mu

fig. 5.22

O

Solution: At some angular displacement θ, the situation isshown in figure 5.23. Applying work energytheorem on the ball, we get,

allw k= ∆

⇒ T g f iw w k k+ = −

θ

mg

VT

v l

2

l

ω

θ

x s

fig. 5.23

cos(1 cos )

x l ll

θ= –= − θ

O

⇒ 2 21 102 2gw mv mu+ = − VT

⊥#

⇒ 2 21 12 2

mg h mv mu− ∆ = −

⇒ 2 21 1.2 2

mg x mv mu− = −

⇒ 2 2 2v u gx= −

⇒ 2 2 2 (1 cos )v u gl θ= − − (i)

⇒ 2 2 (1 cos )v u gl θ= − −

At the position shown in figure 5.23, the ball is moving in a vertical circle of radius l and has a speedv, therefore, its radial acceleration is 2/ .v l Applying Newton’s 2nd law along the radial direction atthis moment, we get

netF ma=

⇒2

cos vT mg ml

θ− =

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⇒2

cos 2 (1 cos )muT mg mgl

θ θ= + − − [using equation (i)]

⇒2

3 cos 2 uT mg mg ml

θ= − +

* You should notice that we had done this problem in the previous chapter also(CIRCULAR MOTION). That time we solved this problem by analyzing the varying tangentialacceleration of the ball.

Example – 7

R

m

smoo

th

fig. 5.24

A small ball is released from the top of a smoothhemispherical surface fixed on a horizontal plane as shownin figure 5.24. If m be the mass of the ball and R be theradius of the hemispherical surface, find the speed of theball and normal contact force between the ball and thehemispherical surface as a function of θ, where θ is theangle between radial position of the ball with respect tothe centre of the hemisphere and the vertical direction.

Solution: The position of the ball when it moves through an angle θ is shown in figure 5.25. As the ball is movingon a circular path of radius R, at the shown position it has a centripetal acceleration of magnitude

2/ .v R Although, I am not discussing about the tangential acceleration of the ball at the shown moment,you should not forget that it is also present.Applying Work Energy Theorem, we get,

allw k= ∆

⇒ N g f kiw w k k+ = −

smoo

th

θ

fig. 5.25

θ 2vR

xs

mgv

m

N

⇒ 210 02

mg h mv− ⋅∆ = −

⇒ 21( )2

mg x mv− − =

⇒ 2 2v gx= ...(i)

⇒ [ (12 cos )]v gx x l θ= −=

Applying Newton’s 2nd law along the radial direction, we get.

netF ma=

⇒ 2

cos 2 (1 cos () [u isin )g ]mvmg N mgl

θ θ− = = −

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⇒ 3 cos 2N mg mgθ= −

* Notice that at 1cos (2/3),θ −= N becomes zero. What is the physical significance of this θ ? Whatwould happen when the ball passes this position?

Example – 8

The kinetic energy of a particle moving along a circle of radius R depends upon the distance covered as 2,k sα=where α is a constant. Find the magnitude of the force acting on the particle as a function of s.

Solution: Let us first find the tangential and normal components of the net force acting upon the particle, then wecan find the net force by adding these two components. If tF be the tangential component of the netforce, then

2( )

tdk d sFds ds

α= =

2 sα=

If nF be the normal component of the net force, then

2

nvF mR

=

22 1

2mv

R =

2 22 ( ) [ ]s k sR

α α= =∵

22 s

Rα=

∴ 2 2net t nF F F= +

( )22

2 22 ssR

αα

= +

2

2 1 ssR

α = +

2 22 s R sRα= +

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Example – 9

A particle of mass m starts moving so that its speed varies according to the law v sα= , where α is a positiveconstant, and s is the distance covered. Find the total work performed by all the forces which are acting on theparticle during the first t seconds after the beginning of motion.

Solution: Using work energy theorem, we have, work done by all forces

(change in K.E.)allw k= ∆

f ik k= −

21 0 [ initially 0 0 0]2

mv s v k= − = ⇒ = ⇒ =∵

212

mv=

212

m sα= ...(i)

So, now we have to find s as a function of t. We have

ds dsv sdt dt

α= ⇒ =

⇒ds dt

sα= ⋅

⇒0 0

s tds dts

α=∫ ∫

⇒ 002

s ts tα=

⇒ 2 s tα=

⇒ 2 2

4ts α= ...(ii)

∴ 2

2allm sw α= [Using (i)]

4 2

8m tα= [Using (ii)]

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Example – 10

A particle of mass m moves along a circle of radius R with a normal acceleration varying with time as 2,na tα=where α is a positive constant. Find the time dependence of the work done by all the forces.

Solution: We have,

2na tα=

⇒2

2v tR

α=

⇒ 2 2v Rtα= ...(i)

∴ Work done by all forces,

allw k= ∆

f ik k= −

21 02

mv= −

212

mv=

⇒2

2allm Rtw α= [Using equation (i)]

Example – 11

A chain of mass m and length l rests on a rough surfaced table so that one of its ends hangs over the edge.The chain starts sliding off the table all by itself provided the overhanging part equals 1/3 of the chain length. Whatwill be the total work performed by the friction forces acting on the chain by the moment it slides completely offthe table?Solution: When the chain has fallen by a distance x its position on the table is shown in figure 5.26. At this

instant friction is opposing the motion of the chain and is acting towards the left. If µ be the frictioncoefficient between the chain and the table and m′ be the mass of the part of the chain on the surfaceof the table, then friction force is

'f m gµ=

23

m l x gl

µ = − l/3

x

2 /3lx

fig 5.26

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23

mg l xl

µ = − �

In the next infinitesimally small time interval dt if the chain slides further by dx, then work done by thefrictional force can be given as

dw f dx= − ⋅

23

mg l x dxl

µ = − ⋅

Therefore, till the moment when the chain leaves the table surface completely (i.e., value of x becomes2l/3), net work done by friction forces is

w dw= ∫2 /3

0

23

x lmg l x dxl

µ = = − − ⋅ ∫

( )22 /32 23 3 2

lmg l ll

µ = − × −

2(2 /3)

2mg ll

µ= −29

mglµ= −

You should note that µ was not given in the question. You have to find it on your own. (Of course youcan find it from the statement of the question).

[Suppose you have to find out the speed of the chain at the moment it just leaves the table, then thatcan be found by using the result above in the work energy theorem. In this case there are only threeforces acting on the chain: gravity, friction and normal contact force from the table. Normal contactforce is not doing work so you need work done by friction and gravity only to find the change in K.E.of the chain, which would lead you to the final speed of the chain. And hence in this way you can avoidcomplicated equations and their solutions which you’d have encountered if you would have chosenmethods learnt in chapter NEWTON’S LAWS OF MOTION.]

Example – 12

In the previous example find the work done by gravity (on the chain) for the same duration.

Solution: For the same time interval dt which we considered in the previous example, work done by gravity onthe chain is

3gm ldw x g dxl

= + ⋅

3mg l x dxl

= + ⋅

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Hence, net work done on the chain by gravity till the moment it leaves the table completely is

g gw dw= ∫2 /3

0 3

x lmg l x dxl

= = + ⋅ ∫

22 (2 3)3 3 2

mg l l ll

= ⋅ +

/ l/3

x

2 /3lx

fig 5.27dx

2 22 29 9

mg l ll

= +

49

mgl=

Note: � In the next topic we will study the concept of CENTRE OF MASS. When you are familiar withthat concept, you find the work done by gravity in a much easier way, although the methoddiscussed here is also simple and easy.

� If v be the speed of the chain when it just leaves the horizontal surface, then applying work energytheorem on the chain between the moments when it started sliding over the horizontal surface andwhen it just left the surface, we get

allw k= ∆

⇒ g f N f iw w w k k+ + = −

⇒24 2 1(0) 0

9 9 2mgl mgl mvµ + − + = −

⇒ 24 2 19 9 2

mgl mgl mvµ− =

Put the value of µ and then solve the above equation to get v.

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Section - 4 POWERPower:The rate of doing work by a force with respect to time is defined as power developed by that force. We know thatwhen a force does work on a body it transfers energy to that body, therefore, power developed by a force is therate of energy transfer by that force. Therefore, for power, P, we can write

dwPdt

= ...(23)

⇒ [ ]dsF F v dw F d sdt

= ⋅ = ⋅ = ⋅!! ! !! !∵P

⇒ ! !P F v= ⋅ ...(24)

Therefore, power developed by a force is the scalar product of the force and the velocity of point of applicationof the force.

The unit of power is Joule/sec which is defined as “watt”.

It is obvious that when force is perpendicular to the velocity, the power developed by the force is zero.

If power developed by a force is known it can be used to calculate the work done by that force. We have,

/P dw dt=⇒ dw P dt= ⋅

⇒ .= =∫ ∫2

1

t

t

w dw P dt ...(25)

Here w is the work done by the force which is developing the power P for the time interval 1 2[ , ].t t

Note: A common unit of power is horsepower.

1 horsepower = 746 watt

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Example – 13

A body of mass m is thrown at an angle α to the horizontal with an initial velocity 0v . Find the mean powerdeveloped by gravity over the whole time of motion of the body, and the instantaneous power of gravity as afunction of time. Assume that the body is thrown at t = 0.

Solution: The average power developed can be defined as the average rate of doing work. Therefore, theaverage power developed by gravity for the time interval [0, t] is .

gwP

t∆

=∆

gF rt

⋅∆=

∆

! !

using rmg v vt

∆ = ⋅ = ∆

!! ! !

is the component of

in the upward directiony

yv v

mg v

= − ⋅

!

is the displacement in the vertical direction

yymgt

∆ ∆= − ⋅ ∆ ( ) (0)

0y t ymg

t−= −−

( )y tmgt

= −

20

1sin2

v t gtmg

t

α ⋅ − = −

0 sin2gtmg v α = − ⋅ −

The average power developed for the entire duration of flight is zero, because for this duration ∆y iszero. Alternatively this could be explained in the following way: as the displacement, ,r∆! is horizontalfor this interval its scalar product with gravity is zero.

The instantaneous power developed by gravity is

gP F v mg v= ⋅ = ⋅! ! ! !

is vertically upward component of y ymg v v v = − ⋅ !

0( sin )mg v gtα= − −

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From the equation above it is clear that while rising, the power developed by gravity is negative andwhile falling, the power developed by gravity is positive. (while rising yv is +ve and while falling yvis –ve.)

Alternate Method: We have

gw mg h∆ = − ∆

∴∴∴∴∴ gg

wP

t∆

=∆

mg ht⋅ ∆= −

∆

hmgt

∆= −∆

Now, this result can be used to obtain the desired result.

Example – 14

For the situation given in example-10 find the dependence of the power developed by all forces on t and find theaverage value of this power for the first t seconds of motion.

Solution: We have,

2na tα=

⇒2

2v tR

α=

⇒ v R tα=

∴ tangential acceleration, tdvadt

=

⇒ ta Rα=

∴ Normal component of net force,

2n nF ma m tα= =

and tangential component of the net force,

.t tF ma m Rα= =

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As we know that work is done by tangential component of the force, power is developed by thiscomponent only. We have,

all netP F v= ⋅! !

( )t nF F v= + ⋅! ! !

⇒ ⋅=all tP F v ( )|| and t nF v F v⊥#! !! !∵ ...(26)

m R R tα α= ⋅

m Rtα=

Again, work done by all forces for the first t seconds of motion can be obtained by either finding theincrement in kinetic energy of the particle or by integrating the power with respect to time. Therefore,for the time interval [0, t], work done by all forces is

0 0

t t

all allw P dt m R t dtα= ⋅ = ⋅∫ ∫

2

2m Rtα=

Therefore, average power developed by all forces for the same interval is

( )2/2work donelength of time interval ( )av

m RtP

tα

= =

2m Rtα=

Alternate-I: We have, ,allP m Rtα= which is a linear function in t, therefore, the average value of allP

final value + initial value2avP =

⇒( ) 02

allav

P tP +=

2m Rtα=

Alternate-II: Using work energy theorem for the time interval [0, t] we get,

all f iw k k k= ∆ = −0

fk=

⇒2

212 2all

m Rtw mv α= =

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∴ Average power developed by all forces

work done by all forceslenth of time interval

=

2( 2)( ) 2

m Rt m Rtt

α α= =/

Alternate-III: Average value of ( )f x over the interval 1 2[ , ]x x is given as2

1

2 1

( )

[ ]

x

xav

f x dxf

x x

⋅

=−

∫

∴ 0 0

( )

( 0)

t t

av

p t dt m R t dtp

t t

α⋅ ⋅= =

−

∫ ∫

2m Rtα=

Alternate-IV: We have tangential acceleration, ,ta Rα= which is constant with time and we also have

initial 0,v = therefore, distance traveled by the particle along the path for the first t seconds is

2 21 12 2ts a t R tα= ⋅ ⋅ =

As the tangential force is constant and work done by normal force is always zero, work done byall forces for the same time interval can be written as

( ) 212all tw F s m R R tα α = ⋅ = ⋅ ⋅

2

2m Rtα=

Therefore, ( )2 2

( )all

m RtP

tα

=/

2m Rtα=

• You should try to understand the physical significance of the equation all tw F s= ⋅ used here.Just give it a thought, and you are sure to get it.

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TRY YOURSELF - II

Q. 1 A heavy stone is thrown from a cliff of height h with a speed v. The stone will hit the ground with maximumspeed if it is thrown(a) vertically downward(b) vertically upward(c) horizontally(d) the speed does not depend on the initial direction.

Q. 2 Total work done on a particle is equal to the change in its kinetic energy(a) always(b) only if the forces acting on it are conservative(c) only if gravitational force alone acts on it(d) only if elastic force alone acts on it.

Q. 3 The kinetic energy of a particle continuously increases with time.(a) The resultant force on the particle must be parallel to the velocity at all instants(b) The resultant force on the particle must be at an angle less than 90° all the time(c) Its height above the ground level must continuously decrease(d) The magnitude of its normal acceleration is increasing continuously.

Q. 4 Under the action of a force, a 2 kg body moves such that its position x as a function of time t is given by3

,3

= tx x is in metre and t in second. Calculate the work done by the force in the first 2 second.

Q. 5 A block shown in figure slides on a semicircular frictionlesstrack. If starts from rest at position A, what is its speedat the point marked B?

B

A

45º1.0m

Q. 6 An object of mass m is tied to a string of length l and a variable force F is applied on it which bringsthe string gradually at angle θ with the vertical. Find the work done by the force F.

θ

F

l

m

Q. 7 A body of mass m accelerates uniformly from rest to v0 in time t0. As a function of t, the instantaneouspower delivered to the body is :

(a) /mv t (b) 2/mv t(c) 2/mvt t (d) 2 2/mv t .

Q. 8 Show that for the same initial speed 0v the speed v of a projectile will be the same at all points at the sameelevation, regardless of the angle of projection.

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Q. 9 A block of mass 10 kg slides down an incline 5 m. long and 3 m high. A man pushes up the block parallelto the incline so that it slides down at constant speed. The coefficient of friction between the block and theincline is 0.1. Find:(a) the work done by the man on the block (b) the work done by gravity on the block(c) the work done by the surface on the block (d) the work done by the resultant forces on the block(e) the change in K.E. of the block.[g = 10 m/s²]

Q. 10 A particle of mass m moves on a straight line with its velocity varying with the distance travelled accordingto the equation ,v a x= where a is a constant. Find the total work done by all the forces during adisplacement from x = 0 to x = d.

Q. 11 Power delivered to a body varies as P = 3 t2. Find out the change in kinetic energy of the body from t =2 to t = 4 sec.

Q. 12 Figure shows a rough horizontal plane which ends in a vertical wall, to which a spring is connected, havinga force constant k. Initially spring is in its relaxed state. A block of mass m starts with an initial velocity utowards the spring from a distance 0l from the end of spring, as shown. When block strikes at the end ofthe spring , it compresses the spring and comes to rest. Find the maximum compression in the spring. Thefriction coefficient between the block and the floor is µ .

l0

km µ

u

Q. 13 A point mass m starts from rest and slides down the surface of a frictionless solid sphere of radius r as infigure. Measure angles from the vertical and potential energy from the top. Find (a) the change in potentialenergy of the mass with angle; (b) the kinetic energy as a function of angle; (c) the radial and tangentialaccelerations as a functions of angle (d) the angle at which the mass flies off the sphere. (e) If there is frictionbetween the mass and the sphere, does the mass fly off at a greater or lesser angle than in part (d)?

r

m

θ

Q. 14 Two disks are connected by a stiff spring. Can one press theupper disk down enough so that when it is released it will springback and raise the lower disk off the table (see figure)? Canmechanical energy be conserved in such a case?

Q. 15 A smooth sphere of radius R is made to translate in a straight line with a constant acceleration a. A particlekept on the top of the sphere is releases from there at zero speed with respect to the sphere. Find thespeed of the particle with respect to the sphere as a function of the angle θ it slides.

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Section - 5 POTENTIAL ENERGY AND ENERGY CONSERVATION

While calculating work done by gravity and an ideal spring, you must have noticed that these were independent ofthe path of motion of the body. Work done by these forces depends only upon initial and final positions. The sameidea leads to the fact that work done by these forces in a closed path is zero. In later chapters you would see thatcoulomb’s forces exhibit similar behaviour. This feature of these forces allow us to group them together in adifferent class of forces which are called conservative forces . Therefore a conservative force can be definedas a force whose work done is always independent of path. Alternatively, we can say that work done by aconservative force depends only upon initial and final positions or we can say that work done by a conservativeforce in a closed path is always zero. You can define a conservative force by any of the three statements. All thesethree statements have the same physical significance. Therefore if any force behaves like this, that force can bedefined as a conservative force. Forces which do not satisfy these conditions can be defined as non conservativeforces.

For a while you are urged not to go into conceptual details of this new topic. We will first develop some methodsbased on the concept of conservative forces. Learn the method and its application initially, and then we will go intothe conceptual details of conservative forces.

The region in which a conservative force is acting is defined as the conservative force field of that force. Forconservative force fields we associate potential energy with them. Why do we do so? This will be clear to you asyou will proceed with this section. In a conservative force field work done by a conservative force is defined asnegative of change in potential energy of the system. If potential energy is denoted by U, then, we define

or= �∆ ∆ = �cons conw U U w ...(27)

Therefore, for gravitational field near the earth’s surface, the change in gravitational potential energy,

( )g gU w mg h∆ = − = − − ∆ [using (6)]

⇒ ∆ = + ∆gU mg h ...(28)

and for a spring and block system,

2 21 12 2sp sp f iU w kx kx ∆ = − = − − −

[using (13)]

⇒ –2 21 1∆ =2 2sp f iU kx kx ...(29)

Now, let us use the definition of change in potential energy in the work energy theorem. We have,

allw k= ∆

⇒ con nonconw w k+ = ∆

⇒ noncon nonconw k w= ∆ −

⇒ ∆ +∆nonconw = k U ...(30)

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33

Sum of kinetic and potential energies is defined as mechanical energy, E. Therefore,

∆nonconw = E ...(31)

i.e., work done by non conservative forces is equal to the change in mechanical energy of the system.

Now, it should be clear that equations (30) or (31) are equivalent to work energy theorem. Instead of using workenergy theorem we can use either of the two equations (30 or 31). So, what’s the advantage if we do this?

The advantage is that if we proceed by this method we are not supposed to calculate the work done by theconservative forces. ∆U on the right side of the equation compensates for that. For any conservative force fieldgenerally we obtain a general expression for potential energy or change in potential energy. Once this result isavailable, we can use equation (30) for all motions in that field. We need not calculate the work done by conservativeforces for all these motions separately. Equation (30) saves us from calculating work done by conservative forceswhich is a must if we follow the work energy theorem method. This is the basic idea behind the development ofthis concept. At this level you may not enjoy or realize the importance of this method because in problems at thislevel you can calculate both work done by conservative forces and change in potential energy very easily. But inthe next level of study of science, in most of the cases you would find the potential energy method much simplerand easier than work energy theorem method, because calculating conservative forces and then their work donewill be really cumbersome and more time consuming. You can consider the example of planetary motions andforces on molecular/atomic levels.

Conservation of mechanical energy:If there are no non conservative forces present in a conservative force field or work done by non conservativeforces is zero, then from equation (30) or (31), we have,

∆ ∆ 0k + U = noncon[when 0]w = ...(32)

or ∆ 0E = ...(33)

i.e., change in mechanical energy of the system is zero. This is known as conservation of mechanical energy.From equation (32) it is clear that if ∆k is +ve then ∆U must be –ve and if ∆k is –ve then ∆U +ve, i.e., increasein kinetic energy is equal to decrease in potential energy and decrease in kinetic energy is equal toincrease in potential energy. The total energy remains the same. If you want a change in the mechanicalenergy of a system, non conservative forces must do work on the system.Let us further extend equation (32). We have

0k U∆ + ∆ = noncon[when 0]w =

⇒ ( )( ) 0f i f ik k U U− + − =

⇒ f f i ik + U = k + U

⇒ f iE = E

Note: While using change in potential energy or mechanical energy in work energy principle you must notconsider work done by conservative forces.

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Now, let us consider the equation (28) once again. We have,

gU mg h∆ = + ∆

⇒ ( )f i f iU U mg h h− = −

⇒ f i f iU U mgh mgh− = − ...(34)

From the equation above we may infer that gravitational potential energy as a function of height (near earthsurface) can be given as

( ) =U h mgh ...(35)

But for the assumption above we need to clarify the following issues:

1. If you assume ( ) ,U h mgh C= + where C is a constant, then also equation (34) holds true;2. From where is the height ‘h’ measured

What is the reference level to measure the height h?The solution of both the problems lies in the fact that the potential energy is not defined absolutely. There is noabsolute value of potential energy. We always define change in potential energy. If we insist on defining potentialenergy then any reference position can be defined as the zero potential energy configuration of the system.Consider equation (34) once again, we have,

f i f iU U mgh mgh− = −

⇒ f ref f refU U mgh mgh− = −

If at refh h= we define 0,refU U= = then

f fU mgh=ref

Reference level from which height is measured is assigned zero value, i.e., 0h =

⇒ ( )U h = mgh

Similarly zero deformation (i.e., x = 0) is defined zero potential energy configuration for the spring and blocksystem. In this case, we get

21( ) =2

U x kx [using equation (29)] ...(36)

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Example – 15

Solve example 2 using potential energy method.

Solution: While the ball slides on the given surface onlytwo forces act on it: (1) gravitational force;(2) normal contact force from the surface.As the gravitational force acting on the ballis conservative in nature, we will not considerthe work done by it. Normal contact forceacting on the ball is the only non conservativeforce acting on the ball. Therefore, fromequation(30), we have

v

N

smoothmg

h

fig. 5.28

nonconw U k= ∆ + ∆

⇒ Nw0 gU k= ∆ + ∆

is always perpendicular

to the direction of motion

N

∵

⇒ 0gU k∆ + ∆ =

⇒20

1( ) 0 02

mg h mv ∆ + − = 0 is the speed of the ball when it arrives

on the horizontal part of the surface

v

⇒ 20

1( ) 02

mg h mv− + =

⇒ 0 2v gh=

Alternate-I: Normal contact force acting the ball is the only non conservative force acting on it and work doneby it zero, therefore,

0U k∆ + ∆ =⇒ k U∆ = −∆i.e., gain in kinetic energy = loss in gravitational potential energy

⇒ 20

12

mv mgh=

⇒ 0 2v gh=

Alternate-II: As the work done by non conservative forces is zero in this case, mechanical energy must beconserved. Therefore

f iE E=

⇒ f f i iU k U k+ = +

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⇒ 2 21 12 2f f i imgh mv mgh mv+ = +

⇒ 20

10 02

mv mgh+ = +

⇒ 0 2v gh=

fig. 5.29

v = 0

O

O

P.E. level or

height level

smoothmg

h

initial position

finalposition

v 0

Example – 16

Solve example 3 using potential energy method.

Solution: Again this example can also be solved in various ways involving the concept of potential energy, as wedid in the last example. As the ball slides down the curved part of the surface, its speed increases withdistance. When the ball arrives on the horizontal part of the surface its speed becomes maximum.When the ball reaches the free end of the spring it begins to compress the spring and its speeddecreases due to the force of the spring acting on it. When speed of the ball falls to zero, it can notcompress the spring further. Hence when compression in spring is maximum, speed of the ball is zero.You should try to analyze the motion of the ball after this moment too.

Therefore from the instant when the ball is released from rest to the moment when compression in thespring is maximum, there are three forces which acted on the ball: (1) weight of the ball, mg; (2)normal contact force, N; (3) spring force. Gravity and spring forces are conservative forces andnormal contact force is the only non conservative force acting on the ball. As the normal contact forceis always perpendicular to the movement of the ball, work done by it is zero.

Method-I: According to equation (30), we have

nonconw U k= ∆ + ∆

⇒ N sp gw U U k= ∆ + ∆ + ∆

⇒ 0sp gU U k∆ + ∆ + ∆ = [ ]0Nw =∵

⇒ [ ]2 2 2 21 1 1 1 02 2 2 2f i f ikx kx mg h mv mv − + ∆ + − =

⇒ 2max

1 0 02

kx mgh− + =max, 0

, 0f i

i f

x x xh h v v

= = ∆ = − = =

⇒ max2mghx

k= .

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Method-II: Here work done by non conservative forces is zero, therefore, total mechanical energy of thesystem remains the same. Hence,

0U k∆ + ∆ =

⇒ 0sp gU U k∆ + ∆ + ∆ =

⇒ 0 0sp g f iU U k k ∆ + ∆ = = = ∵

⇒ sp gU U∆ = −∆

⇒ gain in spring potential energy = loss in gravitational potential energy

⇒ 2max

12

kx mgh=

⇒ max2mghx

k=

Method-III: As the total mechanical energy is conserved in this case,

i fE E=

⇒ , , , ,sp i g i i sp f g f fU U k U U k+ + = + +

⇒ 2max

10 0 0 02

mgh kx+ + = + + 0f ik k= = ∵

⇒ max2mghx

k=

fig. 5.30

initial position(a)

m

k

h

REST

final position(b)

x max

zero gravitational P.E. level

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Example – 17

A string with one end fixed on a rigid wall passing over a fixedfrictionless pulley at a distance of 2m from the wall has a massM = 2 kg attached to it at a distance of 1 m from the wall. Amass m = 0.5 kg attached on the free end is held at rest so thatthe string is horizontal between the wall and the pulley and verticalbeyond the pulley. What will be the speed with which the massM will hit the wall when mass m is released? (g = 9.8 m/s²)

BC

M

m

1 m. 1 m.

fig. 5.31

A

Solution: Here you should notice the following:• Block B moves in a vertical circle with the centre at A when released from rest, as shown in

figure 5.32.

BC

M

m

1 m. 1 m.

fig. 5.32

D

θ

θ

1 m.

A

h

v

u

M

• If block M hits the wall horizontally with speed v, at that instant upward speed of block m iscos .u v θ=

• Gravity (a conservative force) is the only force doing net work on the system “block M + blockm”, because work done by the part of the string between the block m and the vertical wall onethe block m is zero (∵ tension force exerted by this part of the string is always perpendicular tothe movement of the block) and the net work done by the remaining part of the string on the twoblocks is also zero (although work done by it on a single block is nonzero) which can be provedas follows:At any moment power delivered by the part ofthe string between the blocks to the block M isnegative of power delivered by it to the blockm. You can take the example of the moment justbefore block M hits the vertical wall, as shownin figure 5.33. At this moment if v be the speedof the block M and u be the speed of the blockm, as shown in figure, and T be the tension in thethread, then power delivered to block m is

mP T u= + ⋅

C

m

fig. 5.33

Dθ

vuT

T

v cosθM

only tension forces on the blocks from the part of the

string between the blocks are shown

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and that delivered to the block of mass M is

cosmP T v θ= − ⋅

Using constraint equation, we have,

cosu v θ=

∴ m MP P= −

or we can say that the net power delivered to the system “block m and block M” by the string atthis moment is zero. Similarly we can prove this fact for any moment. Hence net work done by thestring on the two blocks during any time interval is zero.Alternate Way:As the string is massless, gain in its kinetic energy must be zero, therefore, net work done on it bythe two blocks must be zero. Hence, net work done by the string on the two blocks is also zero.Now, let us solve for the required unknown v:

Method-I: Let us apply work energy theorem on the system “block m + block M” for the interval starting atthe moment when the system was released from rest and ending at the moment when the block Mis just about to hit the vertical wall. We have,

all f iw k k k= ∆ = −

⇒2 21 1 (0 0)

2 2g Tw w mu MV + = + − +

⇒ 2 21 1( ) 0 ( cos )2 2

Mg AD mg h m v MVθ+ ⋅ − ⋅ + = + ...(i)

From figure 5.32, we have

1 1tan2 2

AD mAC m

θ = = =

⇒2cos ;5

θ =

andincrease in the length of the string on the left side of the pulleyh =

DC BC= −

( ) ( )2 22 1 1 5 1= + − = −

Putting these values in the equation (i), we get,

( ) 2 21 4 12 9 8 1 0 5 9 8 5 1 0 5 22 2 2

V V× ⋅ × − ⋅ × ⋅ × − = × ⋅ × × + × ×

[it is given that m = 0.5 kg and M = 2.0 kg]

Solving the above equation we can find the value of v.

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Method-II: We have,nonconw U k= ∆ + ∆

⇒ Tw U k= ∆ + ∆

⇒ 0U k∆ + ∆ = [ 0]Tw =∵

i.e., mechanical energy of thesystem is conserved. Therefore,we could also use i f i i f fE E U k U k

= ⇒ + = + ⇒ 0m M m MU U k k∆ + ∆ + ∆ + ∆ =

⇒2 21 1( ) ( . ) 0 0 0

2 2mgh Mg AD mu Mv + + − + − + − =

Now, putting the appropriate values as we did in the last method, we can solve for v.

Example – 18 VERTICAL CIRCULAR MOTION

In example 6 we have already calculated the speed of the ball and tension in thethread at some arbitrary angular position θ with respect to the lowermost positionof the ball if the ball has a speed u at its lowermost position. But at that time youmight have thought that what’s the possibility that the ball would complete thevertical circle at all?

mu

fig. 5.34

O

Solution: If the ball is given a horizontal speed u when it is at its lowermost position, as shown in figure 5.34, thefollowing three cases are possible:

(a) The ball completes the vertical circle.

(b) The ball does not reach the horizontal position and oscillates about its initial position, as shown infigure 5.35.

θ θθ

0

0

Ball never reaches the horizontalposition. It oscillates with angularamplitude about its vertical position.

θ

fig. 5.35(c) The ball succeeds in crossing the horizontal position but fails to complete the vertical circle. In this

case we will also discuss whether, when the ball leaves the circle it has zero speed or zero tensionin the string supporting it.Out of these three possible cases, what actually happens would depend upon the value of u. Youmust have an intuition that if u is very small then the ball would oscillate and if u is large than theball would complete the vertical circle.

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Now, consider the results obtained in example 6.At some arbitrary angular position θ speed of the ball, v, is given as

2 2 (1 cos )v u gl θ= − −

⇒ 2 2 2 (1 cos )v u gl θ= − − ...(i)

and tension in the thread, T, is given as2

(3cos 2) muT mgl

θ= − +

⇒ 2 (3cos 2)mT u gll

θ = + − ...(ii)

CASE A: BALL COMPLETES THE VERTICAL CIRCLE:If the ball moves in a complete vertical circle, its distance from the point of suspensionshould be always equal to the length of the inextensible string supporting it and hencestring should always be taut. Therefore,

T > 0

⇒ 2 (3cos 2) 0m u gll

θ + − ≥ [Using equation (ii)]

⇒ 2 (3cos 2) 0u gl θ+ − ≥

⇒ 2 (2 3cos )u gl θ≥ −

⇒ 2 maximum value of (2 3cos )u gl θ≥ −

⇒ 2 5u gl≥at = (topmost position)'2-3cos ' has its maximum value

θ πθ

∵

⇒ ≥ 5u gl

You should not that if 5u gl= , then at the topmostposition T becomes zero but v is nonzero. For

5u gl= at θ = π, equation (i) gives,

v gl=

2/v !T = 0mg

θ=π

= 5u gl

fig. 5.36

and equation (i) gives,

T = 0

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Therefore, for 5 ,u gl= at the topmost position gravity alone is providing thecentripetal acceleration. This fact also gives

2vmg ml

=

⇒ v gl=

∴ velocity of the ball when it reaches the lowermost position is

5u gl= 2 2

using loss in P.E.1 1 (2 ) 52 2

i f f iE E k k

mu mv mg l u gl

= ⇒ = + ⇒ = + ⇒ =

If the speed of the ball at the lowermost position, u, is greater than 5 ,gl then itsspeed at the topmost position is also greater than gl and hence more centripetalforce is required. In this case both tension and gravity contribute to the centripetalforce and hence T > 0.

CASE B: BALL OSCILLATES WITH ANGULAR APTITUDE, θθθθθ0, SMALLERTHAN πππππ/2:

Rearrange equations (i) and (ii) to get2 2( 2 ) 2 cosv u gl gl θ= − + ...(iii)

and2( 2 ) 3 coslT u gl gl

mθ = − +

...(iv)

For this case we have [0, /2).θ π∈ Therefore from equation (iii) and (iv) it is clearthat if 2 2u gl> then neither v nor T becomes zero in this interval of θ. That is neitherthe ball stops nor the string becomes slack in this region. We can also say that if

2 2u gl> then string would definitely cross the horizontal position (i.e., θ = π/2position). Therefore if 2 < 2u gl then the ball can not deflect by π/2 and it wouldoscillate about its lower most position with angular amplitude smaller than π/2. Insuch a case, it is clear from equations (iii) and (iv) that speed of the ball vanishesbefore the tension in the string. That is, in such a case tension in the string is never zeroand speed of the ball is zero at the extreme position (the position from where it startsreturning back towards lowermost position). These facts could also be explained inthe following way:

The position of the ball is shown at some arbitrary angular position θ in figure 5.37(a).If v be the speed of the ball at this position then it is obvious from the figure that at thismoment tension force is balancing the radially outward component of gravity as wellas providing the required centripetal acceleration to the ball and hence it is greaterthan cosmg θ (radially outward component of gravity). Here tangential componentof gravity, sin ,mg θ is retarding the upward (along the circle) motion of the ball asshown in figure 5.37(a). When speed of the ball becomes zero at some angle 0,θthen also tension has to balance the radially outward component of gravity, 0cosmg θ ,

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and hence it can not be zero, as shown in figure 5.37(b). In the same figure youshould notice that, at 0,θ θ= sinmg θ would accelerate the ball towards its lowermost position.

fig. 5.37 ( > cos )

(a)T mg θ

θ

mgsinθ

vT2v

!

mgcosθ

fig. 5.37 ( = cos )

(b)T mg θ

θ

mgcosθ0

mgsinθ0

v=0

T0

If 2 2u gl= then according to equations (iii) and (iv), v and T both vanishsimultaneously at θ = π/2, as shown in figure 5.38. In this case the ball oscillates withangular amplitude π/2.

fig. 5.38(ball just reaches the horizontal position)

T = 0 v = 0

mg

π/2

u gl = 2

CASE C: BALL CROSSES THE HORIZONTAL POSITION BUT DOES NOTCOMPLETE THE VERTICAL CIRCLE.From the previous two cases it must be clear to you that if u is greater than 2glbut smaller than 5gl then the ball would deflect more than /2π but it can notcomplete the vertical circle. Therefore, it would leave the circular path for an angle αgreater than π/2 but smaller than π.

From equations (iii) and (iv) it can be concluded that if2 2u gl> and /2θ π> then T vanishes before v (only if2 5u gl< ), i.e., at some angle α (as mentioned above)

the string becomes slack but the ball still has some nonzerospeed, as shown in figure 5.39. After this particular positiongravity is the only force acting on the moving ball and ithas already left the circular path (because the slack stringmeans that the distance of the ball from the point ofsuspension is smaller than the length of the string),therefore, motion of the ball would be equivalent to thatof a projectile moving in a parabolic path.

T=0

mgα

u

fig. 5.39

v

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ALTERNATE APPROACH:Let us divide the vertical circle under consideration in four quadrants, asshown in figure 5.40. Now, we will analyze the each quadrant separately.

II

I

III

IV

Vertical line

Horizontalline

fig. 5.40

Quadrant I: Some arbitrary position of the ball in this quadrant is shownin figure 5.41. From this figure it is clear that the radialcomponent of the gravity is in the opposite direction of theradial acceleration of the ball and hence in this quadranttension has two roles: first is to balance the radial componentof gravity and second is to provide the necessary centripetalacceleration to the ball, as discussed before. Hence, inthis quadrant tension can never be zero.

θ

mgs inθ

vT2v

!

mgcosθ

ufig. 5.41

If the ball just manages to reach the horizontal position then at this moment alongwith the speed ofthe ball tension in the thread also becomes zero because at this position the radial acceleration andthe radial component of gravity both are zero, as shown in figure 5.38 and hence there is norequirement of tension in the thread. Using work energy theorem or conservation of mechanicalenergy we can prove that for this to happen u should be 2 .gl

Quadrant II: Some arbitrary position of the ball when it is in quadrant II is shown in figure 5.42(a). It is obviousfrom the figure that as the ball moves up, its speed decreases and component of gravity along theradially inward direction increases and hence requirement of tension force becomes less and less asthe ball moves up in this quadrant. Eventually at the highest point of the circle the speed of the ballbecomes minimum and the contribution of gravity in centripetal force becomes maximum and henceat this position requirement of tension is minimum, as shown in figure 5.42(b). Therefore this positioncan be defined as the critical position, because if the ball crosses this position successfully, i.e., ifthe string is taut in this position, it would always be taut or we can say that the ball would completethe vertical circle.

fig. 5.42 (b)

Tmg

2/v !O

v

(requirement of tension is minimum at this position, i.e., the chance of slacking of the string is maximum at this position)

fig. 5.42 (a)

Tmgsin

θ

θ

2/v !

O

mgcosθ

v

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If the ball is just completing the vertical circle we can assume zero tension at the critical position(topmost point) because gravity is there to provide the required centripetal force to the ball. In sucha case if v be the speed of the ball at the highest position, then

2mvmgl

=

⇒ 2v gl=

and if u be the speed of the ball when it reaches the lowermost position, then using

loss in . .f ik k P E= +

we get,

2 21 1 (2 )2 2

mu mv mg l= +

⇒ 2 4u gl gl= +

⇒ 5u gl=

Hence, to complete the vertical circle, ≥ 5u gl .

Note: • To just complete the vertical circle you can not assume zero speed at the highest point. Why so? Tryto answer it on your own.

• The only difference between the analysis of the ball in quadrant II and III is that when the ball ismoving in quadrant II, it is speeding down and while it is moving in quadrant III, it is speeding up.Similar argument can be given for quadrants I and IV.

Summary:

• When 2u gl≤ : The ball oscillates with angular amplitude, 0 /2.θ π≤

• When 2 5gl u gl< < : The ball leaves the vertical circular path at some position (whichwould depend upon u) in the IInd quadrant and thereafter movesin a parabolic path.

• When 5u gl≥ : The ball moves in a complete vertical circular path.

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TRY YOURSELF - III

θ

F

l

m

Q. 1 An object of mass m is tied to a string of length l and avariable force F is applied on it which brings the stringgradually at angle θ with the vertical. Find the workdone by the force F. [Solving using potential energymethod]

Q. 2 A body is dropped from a certain height. When it lost anamount of P.E. ‘U’, it acquires a velocity ‘v’. The massof the body is :(a) 2 / ²U v (b) 22 /v U (c) 2v/U (d) 2/2 .U v

Q. 3 A particle of mass m is attached to a light string of length l, the other end is fixed. Initially the string is kepthorizontal and the particle is given an upward velocity v. The particle is just able to complete a circle.

(a) The string becomes slack when the particle reaches its highest point(b) The velocity of the particle becomes zero at the highest point

(c) The kinetic energy of the ball in initial position was 21 .2

mv mgl=

(d) The particle again passes through the initial position.

Q. 4 A small block of mass m slides along the frictionless loop-the-loop track shown in fig. (a) If it starts from rest at P, what is theresultant force acting on it at Q? (b) At what height above thebottom of the loop should the block be released so that theforce it exerts against the track at the top of the loop is equal toits weight?

R

Q5R

P

Q. 5 A simple pendulum of length l, the mass of whose bob is m, isobserved to have a speed 0v when the cord makes the angle 0θwith the vertical 0(0 /2),θ π< < as in fig. In terms of g and theforegoing given quantities, determine (a) the speed 1v of the bobwhen it is at its lowest position; (b) the least value 2v that 0vcould have if the cord is to achieve a horizontal position duringthe motion.

θ l

m

0

0v

Q. 6 A chain of length l and mass m lies on the surface of a smooth hemisphere of radius R > l with one end tiedto the top of the hemisphere. Find the gravitational potential energy of the chain with respect to the givenP.E. level.

R

ZERO P. E. LEVEL

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Q. 7 A smooth sphere of radius R is made to translate in a straight line with a constant acceleration a. A particlekept on the top of the sphere is released from there at zero speed with respect to the sphere. Find thespeed of the particle with respect to the sphere as a function of the angle θ it slides. [Solve using potentialenergy method]

Q. 8 A block rests on an inclined plane as shown in figure. A spring to which it is attached via a pulley is beingpulled downward with gradually increasing force. The value of sµ is known. Find the potential energy Uof the spring at the moment when the block begins to move.

mk

Fθ

sµ

Q. 9 The particle m in figure is moving in a vertical circle of radius R inside atrack. There is no friction. When m is at its lowest position, its speed is

0v . (a) What is the minimum value mv to 0v for which m will go completelyaround the circle without losing contact with the track? (b) Suppose 0vis 0.775 .mv The particle will move up the track to some point at P atwhich it will lose contact with the track and travel along a path shownroughly by the dashed line. Find the angular position θ of point P.

R

P

v0

m

θ

Q. 10 A chain of length l and mass m lies on the surface of a smooth sphere of radius R > l with one end tied tothe top of the sphere.(a) Find the gravitational potential energy of the chain with reference level at the centre of the sphere.(b) Suppose the chain is released and slides down the sphere. Find the kinetic energy of the chain, when

it has slide through an angle θ.

(c) Find the tangential acceleration dvdt of the chain when the chain starts sliding down.

Q. 11 A spherical ball of mass m is kept at the highest point in thespace between two fixed, concentric spheres A and B (see figure).The smaller sphere A has a radius R and the space between thetwo spheres has a width d. The ball has a diameter very slightlyless than d. All surfaces are frictionless. The ball is given a gentlepush (towards the right in the figure). The angle made by theradius vector of the ball with the upward vertical is denoted by θ(shown in the figure)

θO

R

Sphere B

Sphere A

d

(a) Express the total normal reaction forces exerted by the spheres on the ball as a function of angle θ.

(b) Let AN and BN denote the magnitudes of the normal reaction forces on the ball exerted by thespheres A and B, respectively. Sketch the variations of AN and BN as functions of cosθ in therange 0 θ π≤ ≤ by drawing two separate graphs, taking cosθ on the horizontal axes. Also sketchthe variations of NA and NB as functions of θ.

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Section - 6 MORE ON CONSERVATIVE FORCES

Conservative forces and potential energy againThe interaction of a particle with surrounding bodies can be described in two ways : by means of forces or through theuse of the notion of potential energy. In classical mechanics both ways are extensively used. The first approach,however, is more general because of its applicability to forces in the case of which the potential energy is impossibleto introduce (i.e., non conservative forces). As to the second method, it can be utilized only in the case of conservativeforces.

Our objective is to establish the relationship between potential energy and the force of the conservative field, orputting it more precisely, to define the conservative field of forces ( )F r

! ! from a given potential energy ( )U r! as afunction of a position of a particle in the field.

We have learnt by now that the work performed by conservative forces on a particle during the displacement ofthe particle from one point in the field to another may be described as the decrease of the potential energy of theparticle, that is,

.conw U= −∆

The same can be said about the elementary displacement dr! as well:

condw dU= −

or ⋅ = �! !F dr dU ...(37)

Recalling that ,tdw F dr F ds= ⋅ = ⋅! ! where ds dr= !

is the elementary length covered along the path and tF isthe tangential component of ,F

! we shall rewrite equation (37) as

.tF ds dU⋅ = −

Hence, = −tUFs

∂∂ ...(38)

i.e., the projection of the conservative force at a given point in the direction of the displacement dr! equals thederivative of the potential energy U with respect to a given direction, taken with the opposite sign. The designationof a partial derivative / s∂ ∂ emphasizes the fact of differentiating with respect to a finite direction.

The displacement dr! can be resolved along any direction and, specifically, along the x, y, z coordinate axes. Forexample, if displacement dr! is parallel to the x axis, it may be described as .dr dxi=! The work performed bythe conservative force F

! over the displacement dr! parallel to the x axis is

ˆ( ) ,xF dr F dxi F dx⋅ = ⋅ = ⋅! !!

where xF is the x-component of the force .F!

Substituting the last expression into equation (38), we get

=xUFx

∂−∂ ...(39)

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49

where the partial derivative symbol implies that in the process of differentiating ( , , )U x y z should be consideredas a function of only one variable, x, while all other variables are assumed constant. It is obvious that the equationsfor yF and zF are similar to that for .xF So, having reversed the sign of the partial derivatives of the function Uwith respect to x, y, z, we obtain the components , andx y zF F F of the conservative force .F

!

Hence, we have 2

ˆˆ ˆx yF F i F j F k= + +

!

⇒ ˆ ˆ ˆ = + +

�! U U UF i j k

x y z∂ ∂ ∂∂ ∂ ∂ ...(40)

The quantity in parentheses is referred to as the scalar gradient of the function U and is denoted by grad U or∇∇∇∇∇ U. Generally the second, more convenient, designation where ∇ (“nabla”) signifies the operator

ˆˆ ˆi j kx y z∂ ∂ ∂∇ = + +∂ ∂ ∂

is used. Consequently, we can write,

=!F U∇− ..(41)

i.e., the conservative force F!

is equal to the potential energy gradient, taken with the minus sign. Put simply, theconservative force F

! is equal to the antigradient of potential energy.

Example – 19

The potential energy of a particle in certain conservative field has the following form:

(a) U(x, y) = – αxy, where α is a constant;(b) ( )U r! = ,a r⋅! ! where a! is a constant vector and r! is the position vector of the particle in the field.

Find the conservative field force corresponding to each of these cases.Solution: (a) We have,

ˆ ˆU UF i jx y

∂ ∂= − + ∂ ∂

!

( ) ( )ˆ ˆxy xyi jx y

α α∂ ∂= +∂ ∂

ˆ ˆx yy i x jx y

α α∂ ∂= +∂ ∂

ˆ ˆ( )yi xjα= +

(b) We have, ˆˆ ˆx y za a i a j a k= + +! and

ˆˆ ˆr xi yj zk= + +!

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50

Therefore,U a r= ⋅! !

; thenx y za x a y a z= ⋅ + ⋅ + ⋅

ˆˆ ˆU U UF i j kx y y

∂ ∂ ∂= − + + ∂ ∂ ∂

!

ˆˆ ˆ( )x y za i a j a k= − + +

a= −!

Example – 20

A conservative force F(x) acts on a 1.0 kg particle that moves along the x-axis. The potential energy U(x) is givenas 2( ) ( )U x a x b= + − where a = 20 J, b = 2 m. and x is in meters. At x = 5.0 m the particle has a kinetic energyof 20 J. It is known that there is no other force acting on the system. Based upon this information, answer thefollowing questions:

(a) What is the mechanical energy of the system?

(b) What is the range of x in which the particle can move?

(c) What is the maximum kinetic energy of the particle and the position where it occurs?

(d) What is the equilibrium position of the particle?

Solution: We have,

2( ) ( )U x a x b= + −

220 ( 2)x= + −

We know that at x = 5.0, K.E. = 20 J, therefore, mechanical energy of the particle,

E = Potential energy, U + kinetic energy, k

(at 5) (at 5)U x k x= = + =

220 (5 2) 20= + − +

49 . [Ans.(a)]J=

We know that,

E = U + k

⇒ k E U= −

249 20 ( 2)x = − + − 229 ( 2)x= − −

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As the particle moves on the x-axis, its kinetic energy can never be negative, therefore, we have,

0k ≥

⇒ 229 ( 2) 0x− − ≥

⇒ 2( 2) 29x − ≤

⇒ 2 4 4 29 0x x− + − ≤

⇒ 2 4 25 0x x− − ≤

⇒ [ ]3.38, 7.38x ∈ − [Ans (b)]

When the particle has maximum kinetic energy, we have,

0dkdx

=

⇒229 ( 2)

0d x

dx

− − =

⇒ 0 2 ( 2) (1) 0x− ⋅ − ⋅ =

⇒ 2x =i.e., at 2x = , the particle has maximum kinetic energy which is equal to 29 J. [Ans. (c)]

As the conservative force is antigradient of potential energy, we have,

( ) dUF xdx

= −

220 ( 2)d xdx

+ − = −

[ ]0 2 ( 2) (1)x= − + ⋅ − ⋅

4 2x= − .

When the particle is in equilibrium, net force on it must be zero. As the only force acting on the particleis F(x), in equilibrium position

( ) 0F x =

⇒ 4 2 0x− =

⇒ 2 . [Ans.(d)]x m=

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ALTERNATE METHOD:Let us solve this problem graphically. For that first calculate mechanical energy, E. Following the same procedureas we did in the last method, we get

E = 49J.and we already have,

2( ) 20 ( 2) .U x x= + −The plots of E and U(x) are shown in figure 5.43. From thegraph it is clear that at x = 7.38 and x = –3.38 U becomesequal to E, hence, at these positions k = 0 (because E = k +U). For x > 7.38 and x < –3.38, U become greater than E andhence k would acquire –ve value, which is never possible.Therefore, particle can move only for x greater than –3.38and smaller than 7.38. As the sum of k and U is constant, kwould be maximum when U is minimum. Therefore, at x =2, kis maximum and is equal to 49 – U = 49 – 20 = 29 J. At x = 2,slope of U is zero, therefore, at x = 2, F(x), which negativeof slope of U, is also zero. Therefore, x = 2 is equilibrium

7.382O�3.38

20

49E J=49

U J( )

X( )m

fig. 5.43

position of the particle.

Nature of Equilibrium:Whenever a conservative force is the only force acting on aparticle, equilibrium positions of the particle can be determinedfrom the graph of U when plotted against the position of theparticle.

Consider the case shown in figure 5.44. In this figure potentialenergy, U, of a particle under the action of a conservative force

F(x) is plotted against the position of the particle, x. fig. 5.44

x2O x

U

x1

It is obvious from the graph of U that at 1x x= it has a maximaand at 2x x= it has a minima. Therefore, at both 1x and 2xderivative of U is zero and hence F(x) is zero. Consequently 1xand 2x are equilibrium positions of the particle. Therefore, wecan say that extrema of U occur at equilibrium positions of the particle.

Now, let us analyze the force on the particle when it is in thevicinity of one of its equilibrium positions or it is in the vicinity ofan extre ma of its potential energy.

For values of x very close to 1x but smaller than 1x derivative

F x( ) F x( )x1

xO

U

fig 5.45

of U is positive because its tangent makes an acute angle withthe +ve direction of the x-axis. Therefore, in this region the forceon the particle is along negative x direction (or we can say that itis away from 1x ), as shown in figure 5.45. Similarly, it can be

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justified that for values of x very close to 1x but greater than 1x force, F(x), is positive, i.e., it acts away from 1x ,as shown in figure 5.45. Now, suppose a particle in equilibrium at x = 1x is slightly displaced from this position theeither side and released under the action of conservative force, F(x), only. What would happen now? The forceon the particle would act away from the equilibrium position x = 1x and the particle would never come back tothis equilibrium position. Such an equilibrium position is defined as unstable equilibrium position.

Now, let us analyze the behaviour of F(x) in the vicinity of aminima of U. In figure 5.46 2x x= is a minima of U. Argumentssimilar to what I provided in the previous paragraph lead to thefact that in the vicinity of 2x (a minima of potential energy)conservative force, F(x), acts always towards 2x . Therefore, ifa particle is displaced from 2x on either side and released fromrest and thereafter only conservative force, F(x), acts upon it, itwill return back to its equilibrium position 2( )x x= . (A little morethought would give an idea of oscillation about equilibriumposition, which I don’t want to discuss here.) Such an equilibrium position is defined as a unstable equilibrium position.

F x( ) F x( )x2

xo

y

fig 5.46

Therefore, the position where potential energy, U, Possesses a maxima (first derivative of U is zero and secondderivative is negative) is an unstable equilibrium position and the position where potential energy possesses aminima (first derivative of U is zero and second derivative is positive) is a stable equilibrium position. Therefore,in example 20, x = 2, was a stable equilibrium position.

Example – 21

The potential energy of a particle in a certain field has the form 2/ / ,U a r b r= − where a and b are positiveconstants, r is the distance from the centre of the field. Find:

(a) the value of 0r corresponding to the equilibrium position of the particle, examine whether this position is stable;

(b) the range of the attraction force.Solution: (a) At equilibrium position:

0dUdr

=

⇒2( / / ) 0d a r b rdr

− =

⇒ 3 22 0a br r

− + =

⇒2abr

=

⇒2arb

=

∴ 0 2 /r a b=

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54

Now, ( )2 3 2

2

2a bdd dU drd U r rdr drdr

− + − = =

4 36 2a bdr r

= −

∴ At 0,r r=2

2 4 30 0

6 2d U a bdr r r

= −

4 3

4 36 216 8a b b b

a a⋅ ⋅= −

4 4

3 33 18 4

b ba a

= −

4

318

ba

=

Therefore, at 2

0 2, d Ur rdr

= is positive and hence this is a stable equilibrium position.

(b) Radial component of force, ( ),rF r is antigradient of U with respect to r. therefore,

( )rdUF rdr

= −

3 22a br r

= −

When this force is attractive, it must be along radically inward direction and hence it should benegative. That is,

( ) 0rF r <

⇒ 3 32 0a br r

− <

⇒20a ab b

r r2 − < ⇒ <

⇒2arb

>

Therefore, for 2 , ( )rar F r

b< is repulsive, at

2 , ( )rar F r

b= vanishes and for 2 / , ( )rr a b F r> is

attractive.

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Section - 7 A NOTE ON MASS AND ENERGY (OPTIONAL)

Mass and EnergyOne of the great conservation laws of science has been the law of conservation of matter. From a philosophicalpoint of view an early statement of this general principle was given by the Roman poet Lucretius, a contemporaryof Julius Caesar, in his celebrated work De Rerum Natura. Lucretius wrote “ Things cannot be born fromnothing, cannot when begotten be brought back to nothing.” It was a long time before this concept was establishedas a firm scientific principle. The principal experimental contribution was made by Antoine Lavoisier (1743-1794), regarded by many as the father of modern chemistry. He wrote in 1789 “We must lay it down as inincontestable axiom, that in all the operations of art and nature, nothing is created; and equal quantity of matterexists both before and after the experiment .... and nothing takes place beyond changes and modifications in thecombinations of these elements.:This principle, subsequently called the conservation of mass, proved extremely fruitful in chemistry and physics.Serious doubts as to the general validity of this principle were raised by Albert Einstein in his papers introducingthe theory of relativity. Subsequent experiments on fast moving electrons and on nuclear matter confirmed hisconclusions.Einstein’s findings suggested that, if certain physical laws were to be retained, the mass of a particle had to beredefined as

02 21 /

mmv c

=− . ...(42)

Here 0m is the mass of the particle when at rest with respect to the observer, called the rest mass; m is the massof the particle measured as it moves at a speed v relative to the observer; and c is the speed of light, having aconstant value of approximately 83 10× meters/sec. Experimental checks of this equation can be made, forexample, by deflecting high-speed electrons in magnetic fields and measuring the radii of

0 0.2 0.4 0.6 0.8 1.0

10×10

12×10

14×10

16×10

18×10

�31

�31

�31

�31

�31

,

mkg

v c/

fig 5.47

[ The way an electron’s mass increases as its speedrelative to the observer increases. The solid line is a

plot of m= 2 2 1/20(1 / ) ,m v c −− and the circles are

adapted from experimental values. The curve tendstoward infinity as .v c→ ]

curvature of their path. The paths are circular and the magnetic force a centripetal one 2( / ,F mv r F=and being known).v At ordinary speeds the difference between m and 0m is too small to be detectable. Electrons,however, can be emitted from radioactive nuclei with speeds greater than nine-tenths that of light. In such casesthe results (figure) confirm Equation (42)It is convenient to let the ratio v/c be represented by β. The equation (42) becomes

2 1/20(1 ) .m m β −= −

To find the kinetic energy of a body, we compute the work done by the resultant force in setting the body inmotion. We have,

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56

200

12

vK F dr m v= ⋅ =∫

! !

for kinetic energy, when we assumed a constant mass 0.m Suppose now instead we take into account the variationof mass with speed and use 2 1/2

0(1 )m m β −= − in our previous equation. We find that the kinetic energy is nolonger given by 2

012

m v but instead is2 2 2 2

0 0( ) .K mc m c m m c mc= − = − = ∆ ...(43)

The kinetic energy of a particle is, therefore, the product of 2c and the increase in mass ∆m resulting from themotion.

Now, at small speeds we expect the relativistic result to agree with the classical result. By the binomial theorem wecan expand 2 1/2(1 )β −− as

2 1/2 2 4 61 3 52 8 16

(1 ) 1 ........ .β β β β−− = + + + +

At small speeds / 1v cβ = % so that all terms beyond 2β are negligible. Then

2 2 2 1/20 0( ) (1 ) 1K m m c m c β − = − = − −

( )2 2 2 2 20 0 0

1 1 12 2 2

1 ....... 1 ,m c m c m vβ β= + + − ≅ =

which is the classical result. Notice also that when K equals zero, 0m m= as expected.

The basic idea that energy is equivalent to mass can be extended to include energies other than kinetic. Forexample, when we compress a spring and give it elastic potential energy U, its mass increases from 0m to

20 / .m U c+ When we add heat in amount Q to an object, its mass increases by an amount ∆m, where ∆m is

2/ .Q c We arrive at a principle of equivalence of mass and energy: For every unit of energy E of any kindsupplied to a material object, the mass of the object increases by an amount

2/m E c∆ =This is the famous Einstein formula

2E = ∆mc . ...(44)

In fact, since mass itself is just one form of energy, we can now assert that a body at rest has an energy 20m c by

virtue of its rest mass. This is called its rest energy. If we now consider a closed system, the principle of theconservation of energy, as generalized by Einstein, becomes

20( ) constantm c∑ +∈ =

or 20( ) 0,m c∆ ∑ + ∑∈ =

where 20m c∑ is the total rest energy and ∑∈ is the total energy of all other kinds. As Einstein wrote, “Pre-

relativity physics contains two conservation laws of fundamental importance, namely the law of conservation ofenergy and the law of conservation of mass; these two appear there as completely independent of each other.Through relativity theory they melt together into one principle.”

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Because the factor 2c is so large, we would not expect to be able to detect changes in mass in ordinary mechanicalexperiments. A change in mass of 1 gm would require an energy of 139 10× joules. But when the mass of a particleis quite small to begin with and high energies can be imparted to it, the relative change in mass may be readilynotice able. This is true in nuclear phenomena, and it is in this realm that classical mechanics breaks down andrelativistic mechanics receives its most striking verification.

A beautiful example of exchange of energy between mass and other forms is given by the phenomenon of pairannihilation or pair production. In this phenomenon an electron and a positron, elementary material particlesdiffering only in the sign of their electric charge, can combine to literally disappear. In their place we find high-energy radiation, called γ-radiation, whose radiant energy is exactly equal to the rest mass plus kinetic energies ofthe disappearing particles. The process is reversible, so that a materialization of mass from radiant energy canoccur when a high enough energy γ-ray, under proper conditions, disappears; in its place appears a positron-electron pair whose total energy (rest mass + kinetic) is equal to the radiant energy lost.

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TRY YOURSELF - IV

Q. 1 Show that 2mc has the dimensions of energy.

Q. 2 If the potential energy of a two particle system separated by a distance r is given by ( ) ,AU rr

= where A

is a constant, find the redial force, ,rF that each particle exerts on the other.

Q. 3 The potential energy of a conservative system is given by 2U ax bx= − where a and b are positiveconstants. find the equilibrium position and discuss whether the equilibrium is stable or unstable.

Q. 4 Give physical examples of unstable equilibrium. Of stable equilibrium.

Q. 5 “The position of maximum potential energy is the position of unstable equilibrium”. State whether thestatement is true of false, with short reason.

Q. 6 For the potential energy curve shown in figure

A

B

C

D

6 x m( )

U

842

E

(a) Determine whether the conservative force F ispositive, negative or zero at the five points A, B,C, D and E.

(b) Indicate points of stable and unstable equilibrium.

Q. 7 In the figure shown the potential energy U of a particle isplotted against its position ‘x’ from origin. Then which ofthe following statement is correct. A particle at :

(a) x1 is in stable equilibrium(b) x2 is in stable equilibrium(c) x3 is in stable equilibrium

x

U

x3x2x1O

(d) none of these

Q. 8 The given plot shows the variation of U, the potentialenergy of interaction between two particles with thedistance separating them, r :

(i) B and D are equilibrium points r

A

B

CD

E

F

U

(ii) C is a point of stable equilibrium(iii) The force of interaction between the two particles

is attractive between points C and D and repulsivebetween points D and E on the curve.

(iv) The force of interaction between the particles is repulsive between point E and Fon the curve.

Which of the above statements are correct.(a) (i) and (iii) (b) (i) and (iv)(c) (ii) and (iv) (d) (ii) and (iii)

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Q. 9 A certain peculiar spring is found not to confirm to Hooke’s law. The force (in Newton’s) it exerts whenstretched a distance x (in meters) is found to have magnitude 252.8 38.4x x+ in the direction opposingthe stretch. (a) Compute the total work required to stretch the spring from x = 0.50 to x = 1.00 meter. (b)With one end of the spring fixed, a particle of mass 2.17 kg is attached to the other end of the spring whenit is extended by an amount x = 1.00 meter. If the particle is then released from rest, compute its speed atthe instant the spring has returned to the configuration in which the extension is x = 0.50 meter. (c) Is theforce exerted by the spring conservative or non conservative? Explain.

Q. 10 If the magnitude of the force of attraction between a particle of mass 1m and a mass 2m is given by

1 22

m mF kx

= where k is a constant and x is the distance between the particles, find (a) the potential

energy function and (b) the external work required to increase the separation of the masses from 1x x=to 1 .x x d= +[Assume, U = 0 when x → ∞ ].

Q. 11 The magnitude of the force of attraction between the positively charged nucleus and the negatively charged

electron in the hydrogen atom is given by 2

2eF kr

= where e is the charge of the electron, k is a constant,

and r is the separation between electron and nucleus. Assume that the nucleus is fixed. The electron,initially moving in a circle of radius 1R about the nucleus, jumps suddenly into a circular orbit of smallerradius 2.R

(a) Calculate the change in kinetic energy of the electron, using Newton’s second law.(b) Using the relation between force and potential energy, calculate the change in potential energy of the

atom.(c) Show by how much the total energy of the atom has changed in this process. (The total energy will

prove to have decreased; this energy is given off in the form of radiation.)[Assume, U = 0 when x → ∞ ].

Q. 12 Given below (figure) are examples of some potential energy functions in one dimension. The total energyof the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in whichthe particle cannot be found for the given energy. Also, indicate the minimum total energy the particle musthave in each case. Think of simple physical contexts for which these potential energy shapes are relevant.

(a)

a

E x

U0

O

U x( )

(b)

O x

U3

U x( )

a b c d

U2

U1

U0

E (c)

a

E x

U0

O

U x( )

�U1

b(d)

E

x

U0

U x( )

-b/2

1

-a/2 a/2 - /2b

U

.

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Section - 8 MISCELLANEOUS EXAMPLES

Example – 22

A block of mass 2.0 kg is moved from rest on a rough horizontal surface by applying a force 2( ) 15F x x x= + −from 0 0x = ⋅ to 1 0.x = ⋅ If friction coefficient between the block and horizontal surface be 0 2⋅ , find the gain inkinetic energy of the block. [Take g = 10 m/s²]

Solution: According to the work energy theorem,

k∆ = work done by all forces

as the block is moving on a horizontal surface, work done by gravity and normal contact force from the surface is zero

work done by ( ) work done by frictionF x

= +

2

1

( ) [ 1 0 m.]x

x

F x dx mg d dµ= ⋅ − ⋅ = ⋅∫

12

0

(15 ) 0 2 2 0 10 1 0x x dx= + − ⋅ − ⋅ × ⋅ × × ⋅∫1 12 3

10

0 0

15 42 3x xx= + − −

1 115 42 3

= + − −

11.17= J.

Example – 23

A body of mass m was slowly hauled up the hill shown in figure5.48 by a face F

! which at each point was directed along a

tangent to the trajectory. Find the work performed by this force,if the height of the hill is h, the length of its base l, and the coefficientof friction µ.

l

h

m

F!

fig. 5.48

Solution: An elementary part, of length ds, of the path of thebody and forces acting upon the body when it wason that part of the path are shown in figure 5.49. Theapplied external force F is acting tangentially up thesurface, friction force f is acting tangentially down

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61

the surface, normal contact force N is perpendicular to the surface (and hence is perpendicular to thedirection of motion of the body) and weight, mg , of the body is acting vertically downwards.

f

NF

θ

θ

mg

f

dsN

F

mgsinθmgcosθ

m

fig. 5.49

ds

As the body is being moved slowly, net force on the body should be zero and friction would be kineticin nature. Therefore,

cosF N mgµ µ θ= =

and sin cos sinF f mg mg mgθ µ θ θ= + = +

As the body is moved over this elementary part of the path, from figure 4.9, work done by F is

[ is along ]dw F ds ds F= ⋅ ∵

[ cos sin ]mg mg dsµ θ θ= + ⋅

[ cos sin ]mg ds dsµ θ θ= ⋅ ⋅ + ⋅

Now, if we define the horizontal direction as the x-direction and vertically upward direction as the y-direction, as shown in figure 5.50, then we have

cosds dxθ⋅ =

and sinds dyθ⋅ = θ dy

dx

dsy

x

fig. 5.50

Therefore,

dw mg dx mg dyµ= ⋅ + ⋅

Hence, net work done by force F is

fin fin

in in

.w dw mg dx mg dyµ= = ⋅ +∫ ∫ ∫

⇒ ⋅ ⋅w = µmg l + mg h

Note: As the body is moved slowly or we can say as the equilibrium is always maintained, gain in K.E. of thebody must be zero. Therefore, using work energy theorem, we get,

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0allw k= ∆ =

⇒ 0F N f gw w w w+ + + =

⇒ ( ) 0 0fmgl mgh w mghµ + + + − =

⇒ fw mglµ= −

⇒ work done by friction force = �µmg � l

Now, notice the result obtained above for the work done by the friction force. This work done is equivalentto the work friction would have done if the body was given the same horizontal displacement on a horizontalsurface with the same roughness.

Example – 24

AB is a quarter of a smooth circular track of radius 4m.,r = as shown in figure 5.51. A particle P of mass m = 5kg moves along the track from A to B under the action of the following forces:

(i) A force 1,F always directed towards B, its magnitude isconstant and is equal to 4 newton.

(ii) A force 2,F which always directed along the tangent tothe circular track, its magnitude is (20 – s) newton,where s is the distance traveled.

(iii) A constant horizontal force 3F of magnitude 25 newton.

If the particle starts at A with a speed of 10 m/s, what is the speed at B?

F3

F2F1

r BO

A

r

fig. 5.51

Solution: First of all let us calculate work done by each of thethree forces.Work done by 1 2 and F F : When the particle has alreadymoved a distance s away from A its position is shown infigure 5.52. If the particle has rotated about the centre of thecircular path, point O, by an angle θ and α be the anglebetween OB and direction of 1,F

! then in ,OPB∆ we have,

( ) ( )2π θ α α π − + + =

d lF1

r BO

A

r

fig. 5.52

π θ2θ

α

!!

P

s

⇒ 22πα θ= +

⇒ 4 2π θα = +

Now, if the particle is displaced by dl!

in next time interval dt, work done by 1F!

is

1 1dw F d l= ⋅!!

1 cos2

F dl π α = ⋅ ⋅ −

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63

1 sinF dl α= ⋅ ⋅

1 sinF ds α= ⋅ ⋅ for infinitesimally small displacement ,

(distance moved along path)dl

dl ds

=

!∵ !

1 sin4 2

F r dπ θ θ = + ⋅ ⋅ we have, /s r

s rds r d

θθ

θ

= ⇒ = ⋅ ⇒ = ⋅

16sin4 2

dπ θ θ = + ⋅

In the same time interval ‘dt’, work done by 2F!

is given as

2 2dw F dl= ⋅!!

2 2F dl F dl = ⋅ !!

∵ &

2F ds= ⋅(20 )s ds= − ⋅

∴ Net work done by 1F!

is

1 1w dw= ∫/2

0

16 sin4 2

dπ π θ θ = + ⋅ ∫

/2

0

16 cos(1/2) 4 2

ππ θ− = +

{ }32 cos( /2) cos( /4)π π= − −

16 2 J=

22.63 J=

and net work done by 2F!

is

2 2w dw= ∫/2

0

(20 )r

s dsπ

= − ⋅∫/22

/20

020

2

rr ss

ππ= −

2 220

2 8r rπ π= × −

105.8 J=

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64

Work done by 3

!F : As 3F

! is a constant force, work done by 3F ,

3 3 3displacement along w F F= ×

3 25 4 100F r J J= × = × =

∴ Change in K.E. of particle,

allk w∆ =

⇒ 1 2 3f ik k w w w− = + +

⇒ ( )2 21 1 22.63 105 8 1002 2B Amv mv J− = + ⋅ +

∴2 228 43 100

5Bv × ⋅= +

= 13 83 / .m s⋅

Example – 25

A man of height 0 2h m= is bungee jumping from a platform situated a height25h m= above a lake. One end of an elastic rope is attached to his foot and the

other end is fixed to the platform. He starts falling from rest in a vertical position.

The length and elastic properties of the rope are chosen so that his speed willhave been reduced to zero at the instant when his head reaches the surface of thewater. Ultimately the jumper is hanging from the rope, with his head 8 m abovethe water.

(i) find the unstretched length of the rope.

fig. 5.53(ii) find the maximum speed and acceleration achieved during the jump.

Solution: (i) Let us denote the elastic constant (spring constant) of the rope by k and its unstretched length by0.l The maximum length of the rope is 1 0 23 ,l h h m= − = whilst in equilibrium it is

2 (23 8) 15 .l m m= − = Initially, and at the jumper’s lowest position, the kinetic energy is zero. Ifwe ignore the mass of the rope and assume that the jumper’s centre of mass is half-way up hisbody, we can use conservation of energy to write

21 0

1 ( ) .2

mgh k l l= −

In addition, in equilibrium,

2 0( ).mg k l l= −

Dividing the two equations by each other we obtain a quadratic equation for 0,l2 2 20 1 0 1 2 0 02( ) ( 2 ) 4 221 0,l h l l l hl l l+ − + − = + − =

which gives 0 13 .l m=

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(ii) When the falling jumper attains his highest speed, his acceleration must be zero, and so this mustoccur at the same level as the final equilibrium position 2( ).l l=Again applying the law of conservation of energy,

2 22 0 2 0

1 1 ( ) ( ),2 2

mv k l l mg l h+ − = +

where the ratio m/k is the same as that obtained from the equilibrium condition, namely,

2 0 .m l lk g

−=

Substituting this into the energy equation, shows that the maximum speed of the jumper is1 118 ms 65km h .v − −= ≈ It is easy to see that his maximum acceleration occurs at the lowest

point of the jump. Since the largest extension of the rope (10 m) is five times that at the equilibriumposition (2 m), the greatest tension in the rope is 5 mg. So the highest net force exerted on thejumper is 4mg, and his maximum acceleration is 4g.

Example – 26

There are two conservative fields of forces:

(i) ;F ayi=!

(ii) ˆ .F axi byj= +!

Are these forces conservative?

Solution: Let us find the work performed by each force over the path from a certain point 1 11( , )x y to a certainpoint 2 22( , ) :x y(i) ( )ˆ ˆ ˆ( )dw F dl ayi dxi dyj= ⋅ = ⋅ +

!!

ay dx= ⋅

∴ 2

1

;x

x

w dw a y dx= = ⋅∫ ∫

(ii) ( ) ( )ˆ ˆ ˆ ˆdw F dl axi byj dxi dyj= ⋅ = + ⋅ +!!

ax dx dy dy= ⋅ + ⋅

∴ 2 2

1 1

.x y

x y

w dw a x dx b y dy= = ⋅ +∫ ∫ ∫

In the first case the integral depends on the type of y(x) function, that is, on the shape of the path.Consequently, the first field of force is not a conservative field. In the second case both integralsdo not depend on the shape of the path: they are defined only by the coordinates of the initial andfinal points of the path: the second field of force is a conservative field.

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Example – 27

In a certain conservative field a particle experiences the force ˆ ˆ( )F a yi xj= +!

, where a is a constant, and i andj are unit vectors of the x and y axes respectively. Find the potential energy ( , )U x y of the particle in this field.

Solution: Let us calculate the work performed by the force F!

over the distance from the point O (Fig.) to anarbitrary point ( , ).P x y Taking advantage of thiswork being independent of the shape of the path,we choose one passing through the points OMP andconsisting of two rectilinear sections, then

0

.M P

OPM

w Fdr Fdr= +∫ ∫! !! !

p x y( , )y

j

0i M x( , 0)

fig. 5.54

The first integral is equal to zero since at all points ofthe OM section 0y = and .F dr⊥

! ! Along the section

MPx is constant, therefore, ˆF dr F jdy⋅ = ⋅! !!

yF d y a x d y= = and therefore,

0 .P

OPM

w ax dy axy= + =∫We know that this work must be equal to the decreases in the potential energy, i.e., .OP O Pw U U= −Assuming 0,OU = we obtain ,P OPU w− or

( , ) .U x y axy= −

Another way of finding ( , )U x y is to resort to the total differential of that function:

.U UdU dx dyx y

∂ ∂ = + ∂ ∂

Taking into account that / xU x F ay∂ ∂ = − = − and / ,yU y F ax∂ ∂ = − = − we get

( ) ( ).dU a ydx xdy d axy= − + = −

hence U(x, y) = –axy.

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EXERCISE

OBJECTIVEOBJECTIVEOBJECTIVEOBJECTIVEOBJECTIVE[ LEVEL - I ]

Q. 1 Which of the following can be negative(a) kinetic energy (b) potential energy(c) mechanical energy (d) energy.

Q. 2 Work done by force of friction(a) can be zero (b) can be negative(c) can be positive (d) any of the above.

Q. 3 If force is always perpendicular to motion:(a) speed is constant (b) velocity is constant(c) work done is zero (d) K.E. remains constant.

Q. 4 A block of mass m slides down a smooth vertical circular track. During the motion, the block is in(a) vertical equilibrium (b) horizontal equilibrium(c) radial equilibrium (d) none of these.

Q. 5 Two equal masses are attached to the two ends of a spring of spring constant k. The masses are pulled outsymmetrically to stretch the spring by a length x over its natural length. The work done by the spring oneach mass is

(a) 212

kx (b) – 212

kx

(c) 214

kx (d) – 214

kx .

Q. 6 A small block of mass m is kept on a rough inclined surface of inclination θ fixed in an elevator. Theelevator goes up with a uniform velocity v and the block does not slide on the wedge. The work done bythe force of friction on the block in time t will be (wrt to the ground)(a) zero (b) mgvt cos²θ.(c) mgvt sin θ cos θ (d) mgvt sin2θ.

Q. 7 An elevator is moving upward with an acceleration a (and velocity v) when a man inside the elevator liftsa body of mass m through a height h (in time t). The average power developed by the man is (wrtelevator)

(a) ( )m g a ht+ (b) 1( )( )

2m g a v at+ +

(c) mght

(d) 1( )2

mg v at+ .

Q. 8 A particle is rotated in a vertical circle by connecting it to a string of length l and keeping the other end ofthe string fixed. The minimum speed of the particle when the string is horizontal for which the particle willcomplete the circle is(a) gl (b) 2gl(c) 3gl (d) 5gl

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Q. 9 A block of mass m is moving with a constant acceleration a on a rough horizontal plane. If the coefficientof friction between the block and ground is µ, the power delivered by the external agent after a time t fromthe beginning is equal to :(a) ma²t (b) µmgat(c) µ(a + µg)gt (d) m(a + µg)at.

Q. 10 The block of mass m is pulling, vertically up with constant speed, by applying force P. The free end of thestring is pulled by l meter, the increase in potential energy of the block is :

(a) 2

m g l

(b) mgl

(c) 2 mgl P

m(d) 4

m g l.

Q. 11 A block of mass M is pulled along a horizontal surface by applying a force at an angle θ with thehorizontal. The friction coefficient between the block and the surface is µ. If the block travels at a uniformvelocity, the work done by this applied force during a displacement d of the block is

(a) cos

cos sinµMgd

µθ

θ θ+ (b) cosµMgd

θ

(c) sin

cos sinµMgd

µθ

θ θ+ (d) cos

sinµMgd θ

θ

Q. 12 A spring placed horizontally on a rough horizontal surface is compressed against a block of mass mplaced on the surface so as to store maximum energy in the spring. If the coefficient of friction between theblock and the surface is µ, the potential energy stored in the spring is

(a) ² ² ²2

µ m gk (b)

2 22µm gk

(c) 2 2

2µ m g

k(d)

2 23µ mgk

.

Q. 13 Work done in time t on a body of mass m which is accelerated uniformly from rest to a speed v in time t1as a function of time t is given by

(a) 2

1

12

vm tt (b)

2

1

vm tt (c)

22

1

12

mv tt (d)

22

21

12

vm tt .

Q. 14 A block of mass m moving with a velocity v0 on a smooth horizontal floor collides with a light spring ofstiffness k that is rigidly fixed horizontally with a vertical wall. If the maximum force imparted by the springon the block is F, then:(a) F m∝

(b) F k∝

k

m

m

x

B

v0

(c) F∝ v0

(d) None of these.

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Q. 15 The kinetic energy acquired by a mass m in traveling a certain distance d, starting from rest, under theaction of a constant force is directly proportional to

(a) m (b) independent of m

(c) 1/ m (d) m.

Q. 16 A body is moved along a straight line by a machine delivering constant power. The distance moved by thebody in time t is proportional to(a) 1/ 2t (b) 3/ 4t(c) 3/ 2t (d) 2t .

Q. 17 Potential energy function U(r) corresponding to the central force F= K/r² would be

(a) K r (b) K/r(c) Kr² (d) K/r².

Q. 18 A motor drives a body along a straight line with a constant force. The power P developed by the motormust vary with time t as figure.

(a) P

Y

XtO

(b) P

Y

XtO

(c) P

Y

XtO

(d) P

Y

XtO

.

Q. 19 A particle at rest on a frictionless table is acted on by a horizontal force which is constant in size anddirection. A graph is plotted of the work done on the particle W, against the speed of the particle v. If thereare no frictional forces acting on the particle, how graph will look like.

(a) W

Y

XvO

(b) W

Y

XvO

(c) W

Y

XvO

(d) W

Y

XvO

.

Q. 20 A particle is acted upon by a conservative force ( )ˆ ˆ7 6F i j= −!

N (no other force is acting on the particle).

Under the influence of this force particle moves from (0, 0) to (–3m, 4m), then

(a) work done by the force is 3 J (b) work done by the force is –45 J(c) at (0, 0) speed of the particle must be zero (d) at (0, 0) speed of the particle must not be zero

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OBJECTIVEOBJECTIVEOBJECTIVEOBJECTIVEOBJECTIVE[ LEVEL - II ]

Q. 1 The potential energy of a particle varies with position x according to the relation ( ) 3 4 .U x x x= − Thepoint x = 2 is a point of(a) stable equilibrium (b) unstable equilibrium(c) neutral equilibrium (d) none of the above

Q. 2 The K.E. of a body moving along a straight line varies with time as shown in the figure. The force acting onthe body is

(a) zero(b) constant(c) directly proportional to velocity

KE

t(d) inversely proportional to velocity

Q. 3 A ball is projected vertically upwards with an initial velocity. Which of the following graphs best representsthe K.E. of the ball as a function of time from the instant of projection till it reaches the point of projection?

(a) K.E.

Y

XtO

(b) K.E.

Y

XtO

(c) K.E.

Y

XtO

( d) K.E.

Y

XtO

.

Q. 4 A small spherical ball is suspended through a string of length l. The whole arrangement is inside a vehiclewhich is moving with velocity v. Now suddenly the vehicle stops and ball starts moving along a circularpath. If tension in the string at the highest point is twice the weight of the ball then

(a) 5glυ = (b) 7glυ =(c) velocity of the ball at highest point is gl (d) velocity of the ball at the highest point is 3gl

Q. 5 A particle with total energy E moves in one dimension in a region where the potential energy is U(x). Theacceleration of the particle is zero where-

(a) ( )U x E= (b) ( ) 0U x =

(c) ( ) 0dU x

dx= (d)

( )2

2 0d U x

dx=

Q. 6 A block of mass 1 kg slides down a curved track that is one quadrant of a circle of radius 1 m. Its speedat the bottom is 2 m/s. The work done by the frictional force is (g = 10 m/s2)

(a) –8 J(b) +8J(c) 9 J(d) –9J.

R =1m

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Q. 7 A mass of M kg is suspended by a weightless string. The minimum horizontal force that is required todisplace it until the string makes an angle of 45° with the initial vertical direction is :

(a) ( 2 1)−Mg (b) ( 2 1)+Mg

(c) 2mg (d) 2Mg

.

Q. 8 A ring of mass m can slide over a smooth vertical rod as shown in figure. The ring is connected to a springof force constant k = 4 mg/R, where 2R is the natural length of the spring. The other end of spring is fixedto the ground at a horizontal distance 2R from the base of the rod. If the mass if released at a height 1.5 R,then the velocity of the ring as it reaches the ground is :

(a) gR

(b) 2 gR 1.5 R

2.0 R

m

(c) 2gR

(d) 3gR .

Q. 9 A particle which is constrained to move along the X-axis is subjected to a force in the same directionwhich varies with the distance x of the particle from the origin as f(x) = – kx + ax³. Here k and a arepositive constants. For x > 0, the functional form of the potential energy U(x) of the particle is [JEE]

(a)

U x( )

x (b)

U x( )

x (c)

U x( )

x (d)

U x( )

x .

Q. 10 The potential energy function associated with the force 2ˆ ˆ4 2F xyi x j= +!

is :

(a) 22U x y= − (b) 22 constantU x y= − +

(c) 22 constantU x y= + (d) not defined

Q. 11 An automobile engine of mass M accelerates and a constant power P is applied by the engine. Thedistance x covered in time t is given by

(a)1/ 238

9

=

PtxM (b)

1/ 238 =

PtxM

(c)1/23

9

=

PtxM (d)

1/23 =

PtxM .

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Q. 12 In the system shown in the figure the mass m moves in a circular arc of angular amplitude 60° and radiusr. 4m is stationary. Then

mm AB

4m

60°

(a) the minimum value of coefficient of friction between the mass 4 m and the surface of the table is 0.50(b) the work done by gravitational force on the block m is positive when it moves from A to B(c) the power delivered by the tension when m moves from A to B is zero.(d) the kinetic energy of m in position B equals the work done by gravitational force on the block when it

moves from position A to position B.Q. 13 A particle of mass m moves on the x-axis under the influence of a force of attraction towards the origin O

given by F = –k/x² i . If the particle starts from rest at a distance a from the origin the speed it will attainto reach the origin will be :

(a)1 22k a x

m ax−

(b)

1 22k a xm ax

+

(c)1 2k ax

m a x −

(d)1 2

2m a xk ax

−

.

Q. 14 The potential energy for a force field F! is given by ( ) ( ), cosU x y x y= + . The force acting on a particle

at a position given by coordinates 0,4π

is -

(a) ( )1 ˆ ˆ2

i j− + (b) ( )1 ˆ ˆ2

i j+

(c)1 3ˆ ˆ2 2

i j

+ (d)

1 3ˆ ˆ2 2

i j

−

Q. 15 A machine delivers constant power to a body moving along a straight line. The distance moved in time ‘t’increases by what factor from to t = 2 to t = 8 sec.

(a) 8 (b) 4

(c) 4 2 (c) 2 2

Q. 16 A partical moves in a circular path of radius r and its centripetal acceleration is given by : 2 2ca k rt= here

k is a constant. The power delivered to the particle is

(a) 2 22 m k rπ (b) 2 2m k r t

(c)4 2 5

3m k r t

(d) 0

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Q. 17 If allw kE= ∆ then where does the PE∆ (if any) go?(a) Not accounted in the formulae (b) We should add an extra term(c) Has been taken care of in RHS (d) Has been taken care of in LHS

Q. 18 A stone is thrown off a cliff at a speed v at an angle θ with the horizontal. Theplot of its speed with which if hits the earch as a function of θ is:

v

θ

(a) θ

v

+90º�90º 180º 270º (b)

v

270º+90º 180ºθ

(c)

v

θ(d)

v

θ

Q. 19 A ball is tied to a string and given a horizontal velocity as shown. When will the string slack?

(a)2

1 2cos3

gl ugl

θ − −=

(b)3

1 2cos4

gl ugl

θ − −=

θ

u(c)2

1 2cos3 6

ugl

θ − = −

(d)

21 4cos

3 3ugl

θ − = −

Q. 20 A block (B) is attached to two unstretched springs S1 and S2 with spring constant k and 4k, respectively.The other ends are attached to identical supports M1 and M2 not attached to the walls. The springs andsupport have negligible mass. There is no friction anywhere. The block B is displaced towards wall 1 bya small distance ‘x’ and released. The block returns and moves a maximum distance y towards wall 2.

Displacements x and y are measured with respect to the equilibrium position of the block B. Find .yx

B

2 1

BS2 S1

M2 M12 1

x

x

M1S1S2

M2

(a) 4 (b) 2 (c) 12 (d)

14

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SUBJECTIVE

[ LEVEL - I ]

Q. 1 Non-deformable spheres are kept in a closed packed arrangementas shown. Another identical layer is kept on top such that the centreof each sphere in this layer is vertically above those in the bottomlayer. The radius of each sphere is R and mass is m. Is this the moststable configuration in terms of gravitational potential energy?

Q. 2 A partical makes one complete loop in time ‘t’ and then comes torest. You view it from a space ship that is accelerating downwardswith acceleration ‘a’. What is the change in kinetic energy? Doesthis violate the work-energy theorem’s concept that work done a

m

by all forces is equal to the increment in kinetic energy.

Q. 3 The string is pulled slowly as shown till the block of mass ‘m’ makes as angle of 60º with the vertical.What is the work done by ?F

! Pulley shown in the figure is smooth.

Initial Position

m

R

Fsmooth

Q. 4 A body of 4kg mass is placed on a horizontal surface and experiences a force varying with distance asshown in the graph. Find the speed of the body when the force ceases to act on it.

1 2 3 4 5 6 7 8

10

20

30

F t( )

x mts( )OQ. 5 Find the average frictional force needed to stop a car weighing 500 kg in a distance of 25 m if the initial

speed is 72 km/h.

Q. 6 An object is displaced from position vector 1ˆ ˆ(2 3 )r i j m= +! to 2

ˆ ˆ(4 6 )r i j m= +! under a force2ˆ ˆ(3 2 ) .F x i yj N= +

! Find the work done by this force.

Q. 7 If potential energy of a particle moving along x-axis is given by: 3

4 63xU x

= − +

. Here, U is in joule

and x in metre. Find positions of stable and unstable equilibrium.

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Q. 8 Consider the situation shown in figure. Initially the spring is unstretched when the system is released fromrest. Assuming no friction in the pulley, find the maximum elongation of the spring.

m

k

Q. 9 If the system is released from rest, determine the speeds of both masses after B has moved 1 m. Neglectfriction and the masses of pulleys

B4kg

36 kgA

Stationary Pulley

Q. 10 A block of mass m attached with an ideal spring of forceconstant k is placed on a rough inclined plane havinginclination θ with the horizontal and coefficient of friction

1 tan .2

µ θ= Initially the block is held stationary with thespring in its relaxed state, find the maximum extension in

the spring if the block is released.

m

k

θ

Q. 11 The system shown in figure is in equilibrium. Determinethe acceleration of all the loads immediately after the lowerthread keeping the system in equilibrium has been cut.Assume that the threads are weightless, the mass of thepulley is negligibly small and there is no friction at thepoint of suspension. Also find the elastic potential energystored at this moment in the left string.

k k

m1

m2

m3

m4

Q. 12 A smooth table is placed horizontally and a spring ofunstretched length 0l and force constant k has one endfixed to its center. To the other end of the spring is attacheda mass m which is making n revolutions per secondaround the center . Show that the radius r of this uniformcircular motion is 2 2

0 /( 4 )kl k mnπ− and the tension Tin the spring is 2 2 2 2

04 /( 4 )mkl n k mnπ π− .

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Q. 13 A small block of mass 100 g is pressed against a horizontal springfixed at one end to compress the spring through 5.0 cm (figure).The spring constant is 100 N/m. When released, the block moveshorizontally till it leaves the spring. Where will it hit the ground2 m below the spring ?

Q. 14 A block A of mass m is held at rest on a smooth horizontal floor.A light frictionless, small pulley is fixed at a height of 6 m from thefloor. A light in extensible string of length 16 m , connected withA passes over the pulley and another identical block B is hungfrom the string. Initial height of B is 5 m from the floor as shownin figure. When the system is released from rest, B starts to movevertically downwards and A slides on the floor towards right. (a)If at an instant string makes an angle θ with horizontal, findrelation between speed u of A and v of B. (b) Calculate v whenB strikes the floor.

A

P

B

θ

Q. 15 A object is attached to the lower end of a vertical spring and slowly lowered to its equilibrium position.This stretches the spring by an amount d. If the same object is attached to the same vertical spring butpermitted to fall instead, through what distance does it stretch the spring?

Q. 16 The string in figure has a length l = 4.0 ft. When the ball is released, it will swing down the dotted arc (acircle with center at O). How fast will it be going when it reaches the lowest point in its swing?

l

d

O

nail

Q. 17 The nail in figure is located a distance d below the point of suspension. Show that d must be at least 0.6lif the ball is to swing completely around in a circle centered on the nail.

Q. 18 Consider the situation shown in figure. Mass ofblock A is m and that of block B is 2m. Theforce constant of spring is K. Friction is absenteverywhere. System is released from rest withthe spring unstretched. Find:

(a) the maximum extension of the spring mx

A

B

(b) the speed of block A when the extension in the spring is 2mxx =

(c) net acceleration of block B when extension in the spring is 4mxx = .

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Q. 19 In the figure the block A is released from rest when thespring is in its natural length. For the block B of mass mto leave contact with the ground at some stage whatshould be the minimum mass of the block A?

mB

A

Q. 20 A smooth narrow tube in the form of an arc AB of acircle at centre O and of radius r is fixed so that A isvertically above O and OB is horizontal. Particles P ofmass m and Q of mass 2m with a light inextensible stringof length (πr/2) connecting them are placed insidethe tube with P at A and Q at B and released from rest.Assuming the string remains taut during motion, find thespeed of particles whenP reaches B.

AP

O

r

QB

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[ LEVEL - II ]

Q. 1 There is a fixed frictionless ring and three strings passover it symmetrically as shown. The system is releasedfrom rest when the three strings are in the horizontalplane. What is the work done by gravity after sufficienttime has passed? It is given that M > 3m.Assume there isno friction anywhere and all collisions cause thecolliding bodies to stick together.

120°

Mm m

m

R

l

Q. 2 The spring is initially horizontal and its natural length is0.317 mts. It is free to slide on a cylinder fixed perpen-dicular to the wall, as shown. The joint at A is a spheri-cal joint capable of rotating in all directions. The blockis released and allowed to come to rest. What is thework done on the block by the wall w?

(g = 1 m/s2) The spring constant is 20 N/m. The wall

W

5kg

frictionless cylinder

A

1m

1m

1m

1m

Cylinder is fixed perpendicularto the surface of the wall

has coefficient of friction, 0.1.

Q. 3 You whirl a stone attached to a string in a horizontal circle as shown. What is the net work done by tensionon the stone when the stone is brough to this state from the state of rest in vertical position?g = 1 m/s2, L = 2 mts, m = 3 kg.

m

L

ω = rad/s1 2

Q. 4 In the system, the natural length of each spring is 1 mt and theyare both released from rest at the instant shown. Plot the veloc-ityof the block at point A as a function of a ‘m’. What is its value

m

1m

2 mts

A

k N m=10 / 2as 0 ?m → Is this what you would expect to observe partically?If not, why?

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Q. 5 In the figure shown, dimensions of the upper block arenegligible compared to the lower one which has length‘L’ and is moving as with velocity ‘v’ on a frictionlesssurface. Plot the energy wasted by friction as a functionof ‘µ’ the friction coefficient between the blocks, if theupper block is kept on it jently with zerospeed withrespect to the ground. ( ).M m>>

M

L

µ

m

v

Q. 6 An object is spun in a vertical circle as shown. What is the minimum speed ‘u’ required for it to enterregion - III?

II

I

u

L8

IV

III

Q. 7 A ball of mass ‘m’ is released from angle ‘θ ’ as shown andmeets a peg ‘P’ that is fixed on the wall as shown andmoves in acircle there after. What is the minimum value of θ for it tocomplete the circle if the dimensions of the peg are negligible? L

2

L2

L

MP

θ

Q. 8 A sphere of mass m held at a height 2R between a wedgeof same mass and a rigid wall, is released. Assuming thatall the surfaces are frictionless, find the speed of the spherewhen the sphere hits the ground.

Rm

mα

2R

Q. 9 A ring of mass m = 1.2 kg can slide over a smooth vertical rod. A light string attached to the ring is passingover a smooth fixed pulley at a distance of L = 0.5 m from the rod as shown in the figure. At the other endof the string mass M = 3 kg is attached, lying over a smooth fixed inclined plane of inclination angle 37°.The ring is held in level with the pulley and released. Determine the velocity of the ring when the stringmakes an angle (α = 37°) with horizontal.

37°

37°

M

Lm

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Q. 10 Two bars of masses 1m and 2m connected by a non-deformed light spring rest on a horizontal plane. thecoefficient of friction between the bars and the surface is equal to µ. What minimum constant force has tobe applied in he horizontal direction to the bar of mass 1m in order to shift the other bar?

Q. 11 The potential energy corresponding to a certain two-dimensional force field is given by2 21

2( , ) ( ).U x y k x y= + (a) Derive xF and yF and describe the vector force at each point in terms of its

cartesian coordinates x and y. (b) Derive rF and Fθ and describe the vector force at each point in termsof the polar coordinates r and θ of the point. (c) Can you think of a physical model of such a force?

Q. 12 The system of mass A and B shown in the figure is released from rest with x = 0, determine the velocity ofmass B when x = 3 m.

4m4m

m B

x

Am2√

Q. 13 The potential energy of a 2 kg particle free to move along the x-axis is given by 4 2

( ) 5 = − x xU x Jb b

where b = 1m. Plot this potential energy, identifying the extremum points. Identify the regions whereparticle may be found and its maximum speed. Given that the total mechanical energy is (i) 36; (ii) –4 J.

Q. 14 A particle of mass m is kept on a fixed, smooth sphere of radius R at a position where the radius throughthe particle makes an angle of 30° with the vertical. The particle is released from this position. (a) What isthe force exerted by the sphere on the particle just after the release ? (b) Find the distance travelled by theparticle before it leaves contact with the sphere.

Q. 15 A particle of mass m is kept on the top of a smooth sphere of radius R. It is given a sharp impulse whichimparts it a horizontal speed v. (a) Find the normal force between the sphere and the particle just after theimpulse. (b) What should be the minimum value of v for which the particle does not slip on the sphere ? (c)Assuming the velocity v to be half the minimum calculated in part (b), find the angle made by the radiusthrough the particle with the vertical when it leaves the sphere.

Q. 16 A particle of mass ‘m’ moves along the quarter section of thecircular path whose centre is at the origin. The radius of the circularpath is ‘a’. A force ˆ ˆF yi xj= −

! newton acts on the particle,

where x, y denote the coordinates of position of the particle.Calculate

the work done by this force in taking the particle from point A

y

x

( )x, y

B , a(0 )

O A a, ( 0)

m

(a, 0) to point B(0, a) along the circular path.

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81

Q. 17 Which of the following is/are conservative force (s)? Provide explanation/proof too.

(a) 3 ˆ2F r r=! (b)

5 ˆF rr

= −!

(c)( )

( )3/ 22 2

ˆ ˆ3 xi yjF

x y

+=

+

!(d)

( )( )3/ 22 2

ˆ ˆ3 yi xjF

x y

+=

+

!

(e) ˆF ayi=!

Q. 18 A wedge of mass M with a smooth quarter circular plane, is kepton a rough horizontal surface. A particle of mass m is releasedfrom rest from the top of the wedge as shown in the figure. Whenthe particle slides along the quarter circular plane, it exerts a forceon the wedge. The wedge begins to slide when the particle exertsa maximum horizontal force on it. Find the coefficient of frictionbetween the wedge & the horizontal surface.

M

m

Q. 19 A triangle loop ABC of light string is passed over two smallfrictionless pulleys A and B. AB = BC = AC = 2t. Two massesm and M are attached to the mid point O of AB and point Crespectively. If m is released, m & M cross each other at anypoint P as shown in the figure, then show

that 2 3 5

5mM

−> and 5

2OP t= .

m

M

P

C

BA

O

Q. 20 A loop of mass M with two identical rings of mass 32

M each atits top hangs from a ceiling by an inextensible string. If the ringsgently pushed horizontally in opposite directions, find the angulardistance covered by each ring when the tension in the stringvanishes for once during their motion.

m m

M

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82

TRTRTRTRTRY Y Y Y Y YYYYYOURSELF - IOURSELF - IOURSELF - IOURSELF - IOURSELF - IANSWER

1. (a),(c),(d) 2. 1650 J

3. 212

as bs − + 4. 33.33 10 J−×

5. (d) 6. (a) –10J, (b) +10J7. zero 8. zero

9.2

2

34

mvx

TRTRTRTRTRY Y Y Y Y YYYYYOURSELF - IIOURSELF - IIOURSELF - IIOURSELF - IIOURSELF - IIANSWER

1. (d) 2. (a)3. (b) 4. 16 J5. 3.7 m/s 6. mgl (1– cosθ )7. (b)

9. (a) –260 J 10. 212

ma d

(b) 300 J(c) –40 J(d) zero(e) zero

11. 56 J 12. 1/22 2 2 2

02 4 4 ( 22

mg m g k mglk

µ µ µ µ + − −

13. (a) –mgr (1– cos θ )(b) mgr (1 – cos θ )(c) 2g (1 – cosθ ) ; g sin θ(d) cos–1 (2/3)(e) at a greater angle

14. yes; yes 15. [ ]2 sin (1 cosR a gθ θ+ −

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83

TRTRTRTRTRY Y Y Y Y YYYYYOURSELF - IIIOURSELF - IIIOURSELF - IIIOURSELF - IIIOURSELF - IIIANSWER

1. mgl (1– cosθ )

2. (a)

3. (a), (d)

4. (a) 65 ;mg (b) 3R

5. (a) 20 2 (1 cos )V gl θ+ −

(b) 2 cosgl θ

6.2

sinmgR ll R

7. ( ){ }2 sin 1 cosR a gθ θ+ −

8.2(sin cos )

2smg

kθ µ θ+

9. (a) 5gR

(b) P is above the horizontal by ( )1sin 1/ 3−

10. (a) 2

sinmgR ll R

(b) ( )2

0 0sin sin sinmgRl

θ θ θ θ− + + where 0lR

θ =

(c) ( )01 cosgRl

θ− where 0lR

θ =

11. (a) (3cos 2)N mg θ= −

(b) For 1 2cos : 0,3 BNθ − ≤ =

( )3cos 2AN mg θ= − and

for 1 2cos : 0,3 ANθ − ≥ =

( )2 3cos .BN mg θ= −

PHYSICS LOCUS


84

TRTRTRTRTRY Y Y Y Y YYYYYOURSELF - IVOURSELF - IVOURSELF - IVOURSELF - IVOURSELF - IVANSWER

2. 2/A r

3. (stable)2bxa

=

4.

Stable

Unstable

Vertical

Horizontal

5. True6. (a) A –ve

B –ve C +ve D +ve E zero(b) stable equilibrium : x = 6 unstable equilibrium : x = 2

7. (d)8. (c), (d)9. (a) 31.0 J (b) 5.33 m/s (c) Conservative

10. (a) 1 2( ) if ( ) 0m mU x k Ux

= − ∞ = (b) 1 2

1 1

.( )

km m dx d x+

11. (a) 2

2 1

1 12

keR R

−

(b) 2

2 1

1 1keR R

− −

(c) 2

2 1

1 1 12

keR R

− −

12. (a) ;x a≥

(b) ;x φ∈

(c) and ;x a x b≤ ≥

(d) and 2 2 2 2b a a bx x− < ≤ − ≤ <

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85

OBJECTIVEOBJECTIVEOBJECTIVEOBJECTIVEOBJECTIVEANSWER[ LEVEL - I ]

1. (b),(c),(d) 2. (a),(b),(c),(d)

3. (a),(c),(d) 4. (d)

5. (d) 6. (d)

7. (a) 8. (c)

9. (d) 10. (a)

11. (a) 12. (a)

13. (d) 14. (a),(b),(c)

15. (b) 16. (c)

17. (b) 18. (a)

19. (d) 20. (b),(d)

OBJECTIVEOBJECTIVEOBJECTIVEOBJECTIVEOBJECTIVEANSWER[ LEVEL - II ]

1. (d) 2. (d)

3. (c) 4. (b),(d)

5. (c) 6. (a)

7. (a) 8. (b)

9. (d) 10. (a),(b)

11. (a) 12. (a),(b),(c),(d)

13. (a) 14. (b)

15. (a) 16. (b)

17. (d) 18. (a)

19. (a) 20. (c)

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86

SUBJECTIVESUBJECTIVESUBJECTIVESUBJECTIVESUBJECTIVEANSWER[ LEVEL - I ]

1. NO 2. 2 212

ma t

3. 2mgR

4. 65 /m s

5. 4000 N 6. 83 J

7. Stable equilibrium at x = 2 and unstable equilibrium at x = –2.

8.2mg

k

9. speed of 80 /85

B m s= , speed of 3 80 /2 85

A m s=

10.sinmgk

θ

11. 1 2 3 0;a a a= = =

1 2 3 44

4

;m m m ma gm

+ − −=

22( )Energy = 2

m gk

13. 100 cm

14. (a) cosv u θ= (b) 40 /41

m s

15. 2 d

16. 16 ft/sec.

18. (a) 4mg

k (b) 23mgk

(c) 2g

19. 2m

20.2 (1 )3

grπ+

PHYSICS LOCUS


87

SUBJECTIVESUBJECTIVESUBJECTIVESUBJECTIVESUBJECTIVEANSWER[ LEVEL - II ]

1. 2 2 3Mg R Rl mgl+ −

2. 10.03 Joules

3. 7.50 Joules

4.

V

m

5.

E

µ

mv 2

2

v gL2

2

6.3 32

2gL

+

7.1 7sin 4

2 2θ − = −

8. 2 .singR α

9. 1.2 m/s

10. 21 2

mm gµ +

11. (a) , ;x yF kx F ky F= − = −!

always points radially inward.

12. 3.96 m/s.'

13. 104+10

4�

254

x

U

(i) –3 < x < 3; maximum speed = 6.5 m/s.(ii) E = – 4J is not possible

14. (a)32mg

(b)1 1cos 0.43

63R Rπ− − ⋅ '

15. (a)2mvmg

R−

(b) gR(c) 1cos (7 /12) 54.3− °'

16. Zero

17. (a), (b), (c), (d).

18.3

3 2m

m Mµ =

+

20. ( )1 1cos 3−

w.p.energy final colour

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