chapter 6 – work and energyuregina.ca/~barbi/academic/phys109/2009/notes/lecture-14.pdf ·...

Chapter 6 – Work and Energy

Old assignments (solutions have been posted on

the web)can be picked up in my office

(LB-212)

Chapter 6 (Today)

• Work Done by a Constant Force

• Kinetic Energy, and the Work-Energy Principle

• Potential Energy (next class)• Potential Energy (next class)

Recalling Recalling LastLast LectureLectureRecalling Recalling LastLast LectureLecture

The work done by a constant force is defined as the distance moved multiplied by the component of the force in the direction of displacement :

Work Done by a Constant ForceWork Done by a Constant Force

(6-1)


Problem 6-10 (textbook): What is the minimum work needed to push a 950-kg car 810 m up along a 9.0º incline?

(a) Ignore friction.

(b) Assume the effective coefficient of friction retarding the car is 0.25.


Problem 6-10:

Draw a free-body diagram of the car on the incline. Include a frictional force, but ignore it in part (a) of the problem.

The minimum work will occur when the car is moved at a constant velocity.

y

θθ

y

x

m gr

NFr

f rFr

PFr


Problem 6-10:

(a)

Write Newton’s 2nd law in both the x and y directions, noting that the car is unaccelerated

N Ncos 0 cosyF F mg F mgθ θ= − = → =∑

The work done by in moving the car a distance d along the plane (parallel to ) is given by

P Psin 0 sinxF F mg F mgθ θ= − = → =

∑

∑

PFr

PFr

( ) ( )( )o 2 o 6

P P cos 0 sin 950 kg 9.80 m s 810 m sin 9.0 1.2 10 JW F d mgd θ= = = = ×


Problem 6-10:

(b)

Now include the frictional force, given by:

We still assume that the car is not accelerated. We again write Newton’s 2nd

law for each direction:

� The y-forces are unchanged by the addition of friction, and so we still have

fr NkF Fµ=

� The y-forces are unchanged by the addition of friction, and so we still have

� In the x-direction we have

The work done by in moving the car a distance d along the plane (parallel to ) is given by

N c o sF m g θ=

P fr P frsin 0 sin cos sinx kF F F mg F F mg mg mgθ θ µ θ θ= − − = → = + = +∑

PFr

PFr

( )( )( )( )( )

o

P P

2 o o 6

cos 0 sin cos

950 kg 9.80 m s 810 m sin 9.0 0.25 cos 9.0 3.0 10 J

kW F d mgd θ µ θ= = +

= + = ×

Kinetic Energy, and the Work Energy PrincipleKinetic Energy, and the Work Energy Principle

I mentioned, in the beginning of last lecture, that the concept of energy is very important in physics. Yet, I started discussing about work done by a force.

� What is the connection between these two quantities?

A good definition of energy was introduced by the Einstein’s theory of relativity:

E = mc2

But this is beyond the scope of this course.But this is beyond the scope of this course.

In classical mechanics, we can use a less accurate definition of energy as:

“The ability to do work”

We will define translational kinetic energy and some forms of potential energies , though other forms the energy exist such as nuclear energy, heat energy, etc.


Kinetic Energy

A moving object can do work on a second object. Example:

A car hits another car � the first car applies a force on the second (say, at rest) which consequently undergoes a displacement.

This has the implication that the first car has the ability to do work, consequently it has (or carries) energy.

The energy of motion is called kinetic energy (or KE for short).

� But… how do we measure kinetic energy?


How do we measure kinetic energy?

Consider a rigid body of mass m, say the bus in the figure, moving along a line.

A net force acts on the bus changing its velocity from to over a displacement of magnitude d.

The work done on the bus by the force is given by:

Using Newton’s 2nd law, , we can rewrite the above equation as:Using Newton’s 2nd law, , we can rewrite the above equation as:



We can now use the equation of motion

withx – x0 = dv = v2v0 = v1

Previous slide:

(6-2)



We define translational kinetic energy as:

With this definition, we can rewrite eq. 6.2:

(6-3)

Previous slide:

(6-4)

Where,

We can also write

where

Note: Eq. 6.3 is valid in more general cases even when th e force varies.

(6-6)

(6-7)

(6-4)

(6-5)



Equation 6.6 represents the work-energyprinciple which can be stated as

“ The net work done on a object is equal to the chan ge in the

object’s kinetic energy ”

Previous slide:

object’s kinetic energy ”

The work-energy principle is ONLY valid for the net work done on an object:

� in other words, the work done by the net force applied on the object.


Notes:

Observing the previous equations:

We can conclude the following

1) If the net work done on an object is positive, then ∆KE > 0 implying that there is 1) If the net work done on an object is positive, then ∆KE > 0 implying that there is an increase in the object’s velocity ( v2 > v1 ).

2) If the net work done on an object is negative, then ∆KE < 0 implying that there is a decrease in the object’s velocity ( v2 < v1 ).

Notes:

Because work and kinetic energy can be equated , they must have the same units : kinetic energy is measured in joules . It is also a scalar quantity .

(work done by the hammer on the nail)

(vice-versa)


The change (decrease) in kinetic energy of the hammer is equal to the energy gained by the nail (it starts moving)

or, more generally: The decrease in kinetic energy of the hammer is equal to the work the hammer can do on other objects.


Problem 6-21 (textbook): If the speed of a car is increased by 50%, by what factor will its minimum braking distance be increased, assuming all else is the same? Ignore the driver’s reaction time.

Assume:

1) Only friction acts along the direction of the car’s displacement

2) The driver locks the brakes.


Problem 6-21:

The work needed to stop the car is equal to the change in the car’s kinetic energy. That work comes from the force of friction on the car. Assume the maximum possible frictional force, which results in the minimum braking distance. Thus

Where the static friction is used since the driver locks the brakes. The normal force is equal to the car’s weight if it is on a level surface, and so

fr NsF Fµ=

d = stopping distance

it is on a level surface, and so

In the diagram, the car is traveling to the right.

Since , if the velocity is increases by 50%, or is multiplied by 1.5, then dwill be multiplied by a factor of (1.5)2 , or 2.25.

f r sF m gµ=

2o 2 2 2 11 1 1

fr 2 1 12 2 2 cos180

2s

s

vW KE F d mv mv mgd mv d

gµ

µ= ∆ → = − → − = − → =

mgr

NFr

frFr

2

1d v∝


Problem 6-25 (textbook): A 285-kg load is lifted 22.0 m vertically with an acceleration a = 0.160g by a single cable. Determine

(a) the tension in the cable,

(b) the net work done on the load,

(c) the work done by the cable on the load,

(d) the work done by gravity on the load, and (d) the work done by gravity on the load, and

(e) the final speed of the load assuming it started from rest.


Problem 6-25:

(a)

From the free-body diagram for the load being lifted, write Newton’s 2nd law for the vertical direction, with up being positive.

( )( )T

2 3

T

0.160

1.16 1.16 285 kg 9.80 m s 3.24 10 N

F F m g m a m g

F m g

= − = = →

= = = ×

∑Fnet =

( )( )T

TFr

m gr


Problem 6-25:

(b)

The net work done on the load is found from the net force.

( ) ( )( )( )o 2

net net

3

cos 0 0.160 0.160 285 kg 9.80 m s 22.0 m

9.83 10 J

W F d mg d= = =

= ×

TFr

m gr


Problem 6-25:

(c)

The work done by the cable on the load is

( ) ( )( )( )o 2 4

cable T cos 0 1.160 1.16 285 kg 9.80 m s 22.0 m 7.13 10 JW F d mg d= = = = ×

TFr

m gr


Problem 6-25:

(d)

The work done by gravity on the load is

( ) ( )( )o 2 4

G cos180 285 kg 9.80 m s 22.0 m 6.14 10 JW mgd mgd= = − = − = − ×

TFr

m gr


Problem 6-25:

(e)

Use the work-energy theory to find the final speed, with an initial speed of 0.

( )

2 21 1n e t 2 1 2 12 2

3

2

2 1

2 9 .8 3 1 0 J20 8 .3 1 m s

2 8 5 k gn e t

W K E K E m v m v

Wv v

m

= − = − →

×= + = + =

TFr

m gr

2 8 5 k gm

chapter 6 – work and energyuregina.ca/~barbi/academic/phys109/2009/notes/lecture-14.pdf ·...

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