physics work

5 – WORK, ENERGY AND POWER

5.1 Work

“The product of force and displacement ( in the direction of force ), during which the force is acting, is defined as work.”

When 1 N force is applied on a particle and the resulting displacement of the partic e, in the direction of the force, is 1 m, the work done is defined as 1 J ( joule ). The dimensional formula of work is M1 L2 T 2.

The displacement may not be in the direction of the applied force in all cases. In the figure shown, the displacement, d, makes an angle

with the applied force, F. According to the definition of force,

Work, W = force × displacement in the direction of the force

= F ( d cos ) = ( F cos ) ( d ) = ( the component of force in the direc ion of displacement ) × ( displacement )

( i ) For = π / 2, work W = 0, even if F nd d are both non-zero. In uniform circular motion, the centripetal force acti on a particle is perpendicular to its displacement. Hence, the work done due to centripetal force during such a motion is zero.

( ii ) If < 2, work done is posit ve and is said to be done on the object by the force.

( iii ) If / 2 < θ , work d ne is negative and is said to be done by the object against the force.

5.2 Scalar product of two vectors

The scalar product of two vectors, →→B and A , also known as the dot product, is written by

putting a dot ( ) between the two vectors and is defined as:

→→⋅ B A = l →

A l l →B l cos = A B cos , where θ is the angle between the two vectors.

To obtain the scalar product of →→B and A , they are to be drawn from

a common point, O, with the same magnitudes and directions as shown in the figure.

M is the foot of perpendicular from the

head of →→B to A . OM ( = A cos θ is

the magnitude of projection of →→B on A .

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force in the direc ion of

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force in the direc ion of

2, work W = 0, even if F nd d are both non-zero. In uniform circular

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2, work W = 0, even if F nd d are both non-zero. In uniform circular motion, the centripetal force acti on a particle is perpendicular to its displacement.

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motion, the centripetal force acti on a particle is perpendicular to its displacement. Hence, the work done due to centripetal force during such a motion is zero.

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Hence, the work done due to centripetal force during such a motion is zero.

2, work done is posit ve and is said to be done

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2, work done is posit ve and is said to be done

, work d ne is negative and is said to be done

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, work d ne is negative and is said to be done

5.2 Scalar product

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5.2 Scalar product of two vectors

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of two vectors

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The scalar product of two vectors,

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The scalar product of two vectors, putting a dot (

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putting a dot (

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) between the two vectors and is defined as:

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) between the two vectors and is defined as:

→→

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→→B A

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B A =

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→A

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A l

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Similarly, N is the foot of perpendicular from the head of →→A to B and ON ( = B cos is the

magnitude of projection of →→A on B .

∴→→

⋅ B A = AB cos = B ( A cos ) = ( B ) ( OM )

= ( magnitude of →B ) ( magnitude of projection of

→→B on A )

or→→

⋅ B A = AB cos = A ( B cos ) = ( A ) ( ON )

= ( magnitude of →A ) ( magnitude of projection of

→→A on B )

Thus, scalar product of two vectors is equal to the product of magnitude of one vector with the magnitude of projection of second vector on the direction f the first vector.

The scalar product of vectors is zero if the angle between the vectors = / 2, positive if 0 ≤ < 2 and negative if / ≤ .

Properties of scalar product

( 1 ) Commutative law:

→→

⋅ B A = AB cos θ = B A cos =→→

⋅ AB

Thus, scalar product of two vectors s c mmutative.

( 2 ) Distributive law:

→OP =

→A ,

→OQ = B and

→OR =

→C

as shown in the f gure. Now,

) C B ( A→

+→→

⋅ = l→A l l projection of

→+

→C B on

→A l

= l→A l ( ON ) = l

→A l ( OM + MN )

= l→A l ( OM ) + l

→A l ( MN

= l→A l l proj. of B

→ on

→A l + l

→A l l proj. of C

→ on

→A l

=→→

⋅ B A +→→

⋅ C A

Thus, scalar product of two vectors is distributive with respect to summation.

( 3 ) If→A ll

→B , = 0 ,

→→⋅ BA = A B cos 0 = AB and

→→

⋅ A A = l→A l l

→A l = A2 ∴ l

→A l = AA

Thus, magnitude of a vector is equal to the square root of scalar product of the vector with itself.

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the product of magnitude of one vector with

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the magnitude of projection of second vector on the direction f the first vector.

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The scalar product of vectors is zero if the angle between the vectors

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=

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=→→

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→→⋅

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⋅ A

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AB

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B


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=

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=

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B

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B and

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and

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=

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= l

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l→

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→A

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A l

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l l

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=

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l→

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l

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=

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=

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Thus, scalar product of two vectors is distributive with respect to summation.


( 4 ) If→A

→B , = 90 ∴

→→⋅ BA = A B cos 90 = 0

Thus, the scalar products of two mutually perpendicular vectors is zero.

( 5 ) Scalar products of unit vectors in Cartesian co-ordinate system:

∧

⋅ i i =∧j j =

∧k k = 1 and

∧j i =

∧k j =

∧ik = 0

( 6 ) Scalar product in terms of Cartesian components of vectors:

If →A = A x

∧i + A y j + A z k and B = B x

∧i + B y j + B z

∧k , then

→→

⋅ B A = ( A x∧i + A y j + A z

∧k ) ( B x

∧i + B y j + B z k )

= A x B x + A y B y + A z B z

( 7 )→→

⋅ B A = A B cos

∴ cos =BA B A =

2zB 2yB 2xB z A y A x A

z B z A yB A y x B xA 222

This formula is used to find the angle betwe n two vectors.

5.3 Work done by a variable force

‘A force varying with position or time is known as the variable force.’

To find work done under a variable force, consider a two-dime sional motion of a particle moving along a curved path AB as shown in the figure under the effect of a variable force The curved path is divided into

infinitesimally s all line segments 1∆l , 2∆l , …

, ∆l .

Let→1F , 2F , … , nF be the forces acting on

the particle at line segments, 1∆l , 2∆l , … ,

n∆l respectively. As the line segments are very small, the force over each segment can be considered constant.

The total work as the particle moves from A to B can be obtained as the sum of the work done for different line segments as under.

Total work, W = →

nn2211 ∆ F ∆ F ∆ F lll

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B

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B B

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B B

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B z

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z A

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‘A force varying with position or time

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‘A force varying with position or time

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is known as the variable force.’

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is known as the variable force.’

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To find work done under a variable force,

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To find work done under a variable force, consider a two-dime sional motion of a

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consider a two-dime sional motion of a particle moving along a curved path AB as

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particle moving along a curved path AB as shown in the figure under the effect of a

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shown in the figure under the effect of a variable force The curved path is divided into

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variable force The curved path is divided into

infinitesimally s all line segments

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infinitesimally s all line segments

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= ∑B

A∆ F ili , i = 1, 2, 3, …

If

0 ∆

lim

ll l

, then W = dl B

A F = dl

B

A cos F θ

For one-dimensional motion of the particle along X-axis and force acting in he direction of motion as shown in the figure,

Work, W = =2

1

x

xdx F

B

A0 cos dx F o

where x1 and x2 are the x-coordinates of A and B respectively.

It can be seen from the figure that for a small displacement dx, the force F is almost cons ant and hence work done is Fdx which is the a ea of the strip. Total work for motion of the particle from A to B is the sum of areas of uc st ips which is the area under the curve between x = x1and x = x2.

For a particle moving under the ef ect of a constant force

on a curved path from the point 1r to 2r as shown in

the figure, work don W1 = ∫→

2

1

r

r

dr F = ) r r ( F 12 -

( since, F is constant )

Thus, he work done by the constant force is the scalar product of the force and the displacement vector of the pa ticle

5.4 kinetic energy

“The capacity of a body to do work, by virtue of its motion, is known as the kinetic energy of the body.”

More the speed of the object, more is its kinetic energy. The net force acting on a body produces acceleration or deceleration in its motion thereby changing its velocity and kinetic energy. Also, work is done by the net force on the body while causing its displacement.

The work, W, done by the force, F , on a particle of mass, m, causing acceleration, a and

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and hence work done is Fdx which is the a ea of the strip. Total work for motion of the particle

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the strip. Total work for motion of the particle from A to B is the sum of areas of uc st ips

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from A to B is the sum of areas of uc st ips which is the area under the curve between x = x

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which is the area under the curve between x = x


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on a curved path from the point

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on a curved path from the point

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1

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1r

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the figure, work don W1

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( since,

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displacement, d is given by

W = d F = m d a … … … … … ( 1 )

Using the result, v2 - v02 = 2 d a in the above equation,

W = m

2v v 2

02 - =

21 mv2 -

21 mv0

2 , where v0 and v are the sp eds before and

after the application of force respectively. = K - K0 = change in kinetic energy = ∆ K,

where K0 and K are the initial and final kinetic energies. The unit of kinetic energy is the same as that of work, i.e., joule.

Thus, “the work done by the resultant force on a body, in the absence of dissipative forces, is equal to the change in the kinetic energy of the body.”

This statement is known as the work-energy the rem.

When a particle performs uniform circular motion, the centripetal force acting on it is perpendicular to its tangential instantaneous displacement and hence no work is done by the centripetal force and the speed and kin c en rgy of the particle do not change.

If the body is displaced in the directi n of the force acting on it, work is done by the force on the body and is positive. The inet c energy of the body increases in this case. If the displacement of the body is against t e force acting on it, work is done by the body and is negative. The kinetic energy f th body decreases in this case.

Also, kinetic energy, K = 21 mv2 =

m2vm 22

= m2

p2, where p = linear momentum of the body.

5.5 Potential energy

“The capacity of a body to do work, by virtue of its position in a force field or due to its configuration is known as the potential energy of the body.”

Gra itational potential energy:

The gravitational acceleration, g, due to the Earth’s gravitational force can be taken as constant for heights much smaller as compared to the radius of the Earth.

If a body of mass m moves from height y1 to height y2 in the Earth’s gravitational field,

the force on the body = - mg∧j and its displacement = ( y2 - y1 )

∧j

∴ the work done, W = d F

= - mg∧j ( y2 - y1 )

∧j

= mgy1 - mgy2 … … ( 1 )

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energy the rem.

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energy the rem.

When a particle performs uniform circular motion, the centripetal force acting on it is

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When a particle performs uniform circular motion, the centripetal force acting on it is perpendicular to its tangential inst ntaneous displacem nt and hence no work is done by the

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perpendicular to its tangential instantaneous displacement and hence no work is done by the centripetal force and the speed and kin c en rgy of the particle do not change.

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centripetal force and the speed and kin c en rgy of the particle do not change.

If the body is displaced in the directi n of the force acting on it, work is done by the force

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If the body is displaced in the directi n of the force acting on it, work is done by the force on the body and is positive. The inet c energy of the body increases in this case. If the

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on the body and is positive. The inet c energy of the body increases in this case. If the displacement of the body is against t e force acting on it, work is done by the body and is

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displacement of the body is against t e force acting on it, work is done by the body and is negative. The kinetic energy f th body decreases in this case.

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negative. The kinetic energy f th body decreases in this case.

Also, kinetic energy, K =

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2

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21

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1 mv

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mv2

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2


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“The capacity of a body to do work, by

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Here, the work depends on the initial and final positions of the body and not on the path followed. Also, there is no loss of mechanical energy if friction and air resistance are ignored. In this case, the force is known as conservative force and the field as conservative field.

If v1 and v2 are velocities of the body at heights y1 and y2respectively, then according to work-energy theorem,

The work done, W = 21 mv2

2 -

21 mv1

2 = mgy1 - mgy2 [ from

equation ( 1 ) above ] Here, mgy1 and mgy2 are the potential energies of the body at heights y1 and y2 respectively from the surface of the Earth due to the Earth’s gravitational field. Surface of the Earth is the reference level from her potential energy is calculated. Besides, the Earth’s surface, any other level can be chosen as reference level.

Thus, gravitational potential energy of a body of mass m at height h from the surface of the Earth is U = mgh.

From the above equation, 21 mv1

2 + mgy1 = 21 mv2

2 + mgy2

Thus, in a conservative field, he mechanical energy, E, which is the sum of kinetic energy

( K =21 mv2 ) and potent al energy ( U = mgh ) remains constant.

∴ E = K + U.

Thus, the mechani al energy of an isolated system of bodies is conserved in a conservative field. This statement is known as the law of conservation of mechanical energy.

5.6 Elastic potential energy: ( Potential energy due to the configuration of a system )

On end of an elastic spring of negligible mass and obeying Hooke’s law is tied with a rigid wall and to the other end a block of mass m is tied. The block is at x = 0 when the spring is not extended or compressed. On displacing the block, the restoring force is produced in the spring which tries to restore the block to its original position.

The restoring force is directly proportional to the displacement of the block and is in a direction opposite to the displacement.

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of the Earth is the reference level from her potential energy is calculated. Besides, the

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of the Earth is the reference level from her potential energy is calculated. Besides, the Earth’s surface, any other level can be chosen as reference level.

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Earth’s surface, any other level can be chosen as reference level.

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Thus, gravitational potential energy of a body of mass m at height h from the surface of the

From the above equation,

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From the above equation,

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1 mv

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+ mgy

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) and potent al energy ( U = mgh ) remains constant.

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) and potent al energy ( U = mgh ) remains constant.

E = K + U.

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E = K + U.

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Thus, the mechani al energy of an isolated system of bodies is conserved in a conservative

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field. This statement is known as the law of conservation of mechanical energy.

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field. This statement is known as the law of conservation of mechanical energy.

5.6 Elastic potential energy:

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5.6 Elastic potential energy:

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due to the configuration of a system )

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due to the configuration of a system )

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∴ F - x or F = - kx,

where k is the force constant of the spring which is defined as the force required to pull or compress the spring by unit displacement. Its unit is N m and its dimensional formula is M1 L0 T 2.

Work done by the applied force on the spring is,

W = x

0kxdx = k

x

0xdx = k

x

0

2

2x =

21 kx2

This work done on the spring is stored in the form of elastic potential energy of the spring. Taking the potential energy to be zero for x = 0, for hange i length equal to x, the potential stored in the spring will be

U = 21 kx2

Relation between force and potential energy

The change in kinetic energy, K, of the par icle is equal to work, W, done on it by force, F, in displacing it through a small distance, x.

∴ ∆K = W = F ∆x

But, by the law of conservation of mechanical energy,

∆K + U = 0 F x + U 0 ∴ F = x∆U∆ -

∴ instantaneous value of F = -x∆U∆lim

0 x∆ = -

dxdU This equation holds only in the

case of conservative forces.

In the case of a spring, the potential energy, U = 21 kx2

∴ F = -dxdU = -

21 k ( 2x ) = - kx

5.7 Power

“Th time rate of doing work is known as power.” or “Power is defined as the work done per unit time.”

If ∆W is the work done in time interval ∆t, the average power in time interval ∆t is

< P > = t∆

W∆ and instantaneous power at time t is

P = t∆

W∆ lim0 t∆ →

=dt

dW

If dW is the work done by a force F during the displacement dr , then

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omrm of elastic potential energy of the spring. = 0, for hange i length equal to x, the

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K, of the par icle is equal to work, W, done on it by force,

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K, of the par icle is equal to work, W, done on it by force, F, in displacing it through a small distance,

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F, in displacing it through a small distance,

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x.

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x.

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But, by the law of conservation of mechanical energy,

x +

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x +

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U 0

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U 0

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instantaneous value of F =

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In the case of a spring, the potential energy, U =

F =

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F = -

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-

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dx

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dxdU

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dU

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=

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=

5.7 Power

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5.7 Power

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P = dt

dW =→F

dt

dr =

→vF , where v is the instantaneous velocity at time t.

Power is a scalar quantity like work and energy. Its unit is watt ( W ). 1 W = 1 J /s. Its dimensional formula is M1 L2 T 3.

1 kilowatt ( kW ) = 103 watt ( W ) and 1 megawatt ( MW ) = 106 W.

In British system, unit of power is horsepower ( hp ). 1 hp = 746 W.

1 kilowatt-hour (kWh) is the electrical unit of energy (work) as the p oduct f power and time.

1 unit of electrical energy = 1 kWh = 3.6 × 106 J.

5.8 Elastic and inelastic collisions

• In elastic collision, total linear momentum and tot l kinetic energy of the colliding bodies are conserved.

• In inelastic collision, total linear momentum of the colliding bodies is conserved, but part or whole of the kinetic energy is lost in her forms of energy.

• In both, elastic as well as inelastic coll sions, total energy ( energy of all forms, mechanical, internal, sound, etc. ) and total linear momentum are conserved.

Inelastic collision in one dimension:

Partly inelastic collision:

Suppose a sphere A f ma s m1 moving with velocity v1 in X-direction collides with another sphere of mass m2 moving in the same direction with velocity v2. ( v1 > v2 )

Let v1 and v2’ be their velocities in the same ( X ) direction after the collision which we w nt to find.

A cording to the law of conservation of linear momentum,

m1v1 + m2v2 = m1v1’ + m2v2’

or, m1 ( v1 - v1’ ) = m2 ( v2 - v2’ ) … … ( 1 )

Now, a parameter known as coefficient of restitution, e, is defined as under:

Co-efficient of restitution, e = 21

'1

'2

v vv v

- - =

collision the before approach of velocitycollision the after separation of velocity

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om1 kilowatt-hour (kWh) is the electrical unit of energy (work) as the p oduct f power and time.

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om1 kilowatt-hour (kWh) is the electrical unit of energy (work) as the p oduct f power and time.

In elastic collision, total linear momentum and tot l kinetic energy of the colliding bodies

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In inelastic collision, total linear momentum

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In inelastic collision, total linear momentum of the colliding bodies is

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of the colliding bodies isor whole of the kinetic energy is lost in her forms of energy.

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or whole of the kinetic energy is lost in her forms of energy.

In both, elastic as well as inelastic coll s

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In both, elastic as well as inelastic coll smechanical, internal, sound, etc. ) and total linear momentum are conserved.

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mechanical, internal, sound, etc. ) and total linear momentum are conserved.


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Suppose a sphere A f ma s m

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Suppose a sphere A f ma s manother sphere of mass m

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another sphere of mass m2

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2 moving in the same direction with velocity v

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moving in the same direction with velocity v

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For partly inelastic collision, 0 < e < 1. ∴ '

1'

2 v v = e ( 21 v v ) … … … … … ( 2 )

Solving equations ( 1 ) and ( 2 ) for v1’ and v2’, we get

'1v = 2

212

121

21 v m mm ) e 1 (

v m m

em m and 'v2 = 22121

1211 v

m mm em v

m mm ) e 1 (

Thus, v1’ and v2’ can be calculated from the above equations.

Completely inelastic collision:

Putting e = 0 for completely inelastic collision in the above equations,

'1v = 'v2 =

212211

m mvm vm = v ( the common velocity the spheres after collision )

Thus, in completely inelastic collision, the colliding bodies move jointly with a common velocity.

Here, the kinetic energy before collision, Ki = 11vm 2

+ 222vm

21 and

the kinetic energy after collision, Kf = 22 v ) m m (

21 =

2

2122 11

21 m mvm vm ) m m (

21

∴ Kf - Ki = -) m m ( 2) v v ( mm

1

22121 -

The negative sign indicates that the k netic energy decreases during inelastic collision.

Elastic collision in one d mension:

Putting e = 1 for ela tic c llision in equation ( 2 ), we get '1

'2 v v - = ( 21 v v - )

∴ v1 + v1’ = v2 + v2’ … … … … … ( 3 )

Multiplying equations ( 1 ) and ( 3 ), we get

m1 ( v 2 - v1’2 ) = m2 ( v22 - v2’2 ). Multiplying by

21 and rearranging,

21 m1v1

2 +21 m2v2

2 =21 m1v1’2 +

21 m2v2’2

This shows that the kinetic energy is conserved for e = 1. Hence, elastic collision can also be defined as the one in which e = 1.

Special cases of elastic collision:

Taking e = 1 for elastic collision, the velocities of the spheres after the collision are given by

'1v = 2

212

12121 v

m mm 2

v m mm m - and 'v2 = 2

2121

121

1 v m mm m v

m mm 2 - -

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21

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m m

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m em

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m m m m

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= v ( the common velocity the spheres after collision )

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Thus, in completely inelastic collision, the colliding bodies move jointly with a common

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=

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=

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11

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11vm

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vm vm

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vm 11vm 11

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11vm 112

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2

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+

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+

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2

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2 v ) m m (

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v ) m m ( v ) m m (

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v ) m m ( 2 v ) m m ( 2

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2 v ) m m ( 22

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2

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1

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1 v ) m m ( v ) m m (

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v ) m m ( v ) m m ( 2 v ) m m ( 22 v ) m m ( 2

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2 v ) m m ( 22 v ) m m ( 2 v ) m m ( v ) m m (

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v ) m m ( v ) m m ( 2 v ) m m ( 22 v ) m m ( 2

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2 v ) m m ( 22 v ) m m ( 2 v ) m m ( v ) m m (

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v ) m m ( v ) m m ( 2 v ) m m ( 22 v ) m m ( 2

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2 v ) m m ( 22 v ) m m ( 2

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) m m ( 2

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) m m ( 2) v v ( mm

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) v v ( mm1

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1 ) m m ( 2 1 ) m m ( 2

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) m m ( 2 1 ) m m ( 2

2

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22121

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2121 ) v v ( mm 2121 ) v v ( mm

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) v v ( mm 2121 ) v v ( mm


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Putting e = 1 for ela tic c llision in equation ( 2 ), we get

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Putting e = 1 for ela tic c llision in equation ( 2 ), we get

1

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1’ = v

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’ = v2

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2 + v

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+ v


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m

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m1

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1 ( v

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( v

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be defined as the one in which e = 1.


( 1 ) For m1 >> m2, neglecting m2 as compared to m1 in the above equations, we get

v1’ v1 and v2’ 2v1 - v2

This shows that the larger sphere continues to move with the same velocity, whereas the velocity of the smaller sphere increases. If the smaller sphere was at rest, it moves with twice the velocity of the larger sphere after the collision.

( 2 ) For m2 >> m1, neglecting m1 as compared to m2 in the above equatio s, we get v2’ ≈ v2 and v1’ ≈ 2v2 - v1 In this case also, the larger sphere continues to move with he same velocity, but the

velocity of the smaller sphere changes.

( i ) If the smaller sphere were moving with twice he speed of the larger sphere, it becomes stationary after the collision.

( ii ) If the smaller sphere were moving with velocity less than twice the velocity of the larger sphere, it continues to move in he same direction, but with decreased speed.

( iii ) If the smaller sphere were moving with velocity more than twice the velocity of the larger sphere, it rebounds and starts moving in the opposite direction with the velocity given as above.

If the larger sphere was stati nary, it remains stationary. In this case, the smaller sphere rebounds and move with the same speed in the reverse direction.

Summarizing the above t o cases, when one of the two spheres colliding is much more massive than the other then he velocity of the larger sphere remains almost the same after the collision, whereas the velocity of the smaller sphere after the collision is almost twice the velocity of larger sphere less the velocity of the smaller sphere before the collision.

( 3 ) If m1 = m2 then v1’ = v2 and v2’ = v1. Thus, in elastic collision of two spheres of equal mass their velocities get exchanged.

Elastic collisi n in two dimensions

Suppose a sphere of mass m1 moving in

X direction with velocity →1v collides elastically with

a stationary ( →2v = 0 ) sphere of mass m2 as

shown in the figure. After the collision, they move in the directions making angles θ1 and 2 with the

X-axis with velocities '1v and '

2v respectively.

According to the law of conservation of momentum,

m1→1v = m1

'1v + m2

'2v

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om This shows that the larger sphere continues to move with the same velocity, whereas

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in the above equatio s, we get

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In this case also, the larger sphere continues to move with he same velocity, but the

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( i ) If the smaller sphere were moving with twice he speed of the larger sphere, it

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( ii ) If the smaller sphere were moving with velocity less than twice the velocity of the

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larger sphere, it continues to move in he same direction, but with decreased

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( iii ) If the smaller sphere were moving

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( iii ) If the smaller sphere were moving with velocity more than twice the velocity of the

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with velocity more than twice the velocity of the larger sphere, it rebounds and starts moving in the opposite direction with the

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larger sphere, it rebounds and starts moving in the opposite direction with the

If the larger sphere was stat

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If the larger sphere was stati nary, it remains stationary.

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i nary, it remains stationary. sphere rebounds and move with the same speed in the reverse direction.

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sphere rebounds and move with the same speed in the reverse direction.

Summarizing the above t o cases, when one of

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Summarizing the above t o cases, when one of

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massive than the other then he velocity of the larger sphere remains almost the same after

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massive than the other then he velocity of the larger sphere remains almost the same after

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the collision, whereas the velocity of the smaller sphere after the collision

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the collision, whereas the velocity of the smaller sphere after the collision

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velocity of larger sphere less the velocity of the smaller sphere before the collision.

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velocity of larger sphere less the velocity of the smaller sphere before the collision.

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= m

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= m2

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2 then v

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then v

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equal mass their velocities get exchanged.

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equal mass their velocities get exchanged.


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Suppose a sphere of mass m

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shown in the figure. After the collision, they move


Equating the X- and Y-components of the momenta

m1v1 = m1'

1v cos 1 + m2'

2v cos 2 … ( 1 ) and 0 = m1'

1v sin 1 - m2'

2v sin 2 … ( 2 )

As the collision is elastic, kinetic energy is conserved,

∴21 m1v1

2 = 21 m1v1’2 +

21 m2v2’2 … ( 3 )

Using the above three equations, any three unknown quantities can be determined.

5.9 Different forms of energy

Besides mechanical energy, some examples of other forms of nergy are as under.

Internal Energy:

The sum of vibrational kinetic energy and potential energy due to mutual attraction and repulsion between constituent particles of a substa ce is known as internal energy of the body. Due to work done against friction, interna energy and hence temperature of the body increases.

Heat or thermal energy:

The kinetic energy of the constituen particles of a body due to their random motion is known as heat or thermal energy of th body.

The difference between internal energy and heat is analogous to mechanical energy and work. When work is done by body A o body B, mechanical energy of body A decreases and that of body B increases. Similar y, when heat is transferred from body A to body B, internal energy of body A decr ases and that of body B increases.

Chemical energy:

A stable compound has energy less than its constituent elements in free state. This difference is known as c emical energy or chemical binding energy. In a chemical reaction, if the chemical energy of the products is less than that of the reactants, heat is evolved and the reaction is called exothermic. Here, chemical energy got converted into heat energy. Similarly, if the chemical energy of the products is more than that of the reactants, heat is absorbed and the rea tion is called endothermic. Here, heat energy got converted into chemical energy.

Elec rical energy:

The energy associated with electric current is known as electrical energy. When electric current is used in a heater, electrical energy is converted into heat energy. Similarly, in a lamp, it is converted into heat and light energy.

Nuclear energy:

The mass of a nucleus is less than the sum of the masses of its constituent protons and neutrons in free state. The energy equivalent to this mass difference is known as nuclear energy or nuclear binding energy. In a nuclear fission reaction, when heavy nuclei like uranium are bombarded by neutrons, they break up into smaller nuclei releasing huge amount of nuclear energy. Such reactions are used in nuclear reactors for producing power and in

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Besides mechanical energy, some examples of other forms of nergy are as under.

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The sum of vibrational kinetic energy and potential energy due to mutual attraction and

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omrepulsion between constituent particles of a substa ce is known as internal energy of the body. Due to work done against friction, interna energy and hence temperature of the body

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body. Due to work done against friction, interna energy and hence temperature of the body

The kinetic energy of the constituen particles of a body due to their random motion is

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The kinetic energy of the constituen particles of a body due to their random motion is known as heat or thermal energy of th body.

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known as heat or thermal energy of th body.

The difference between internal energy and heat

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The difference between internal energy and heat When work is done by body A o body B, mechanical energy of body A decreases and that

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When work is done by body A o body B, mechanical energy of body A decreases and that of body B increases. Similar y, when heat is

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of body B increases. Similar y, when heat is energy of body A decr ases and that of body B increases.

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energy of body A decr ases and that of body B increases.

A stable compound has energy less than its consti

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A stable compound has energy less than its constiis known as c emical energy or chemical bindi

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is known as c emical energy or chemical bindichemical energy of the products is less than th

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chemical energy of the products is less than th

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reaction is called exothermic. Here, chemical energy got converted into heat energy. Similarly,

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reaction is called exothermic. Here, chemical energy got converted into heat energy. Similarly, if the chemical energy of the products is more than that of the reactants, heat is absorbed

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if the chemical energy of the products is more than that of the reactants, heat is absorbed and the rea tion is called endothermic. Here, heat energy got converted into chemical energy.

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and the rea tion is called endothermic. Here, heat energy got converted into chemical energy.

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The energy associated with electric current is known as electrical energy. When electric www.exam

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The energy associated with electric current is known as electrical energy. When electric www.exam

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current is used in a heater, electrical energy iswww.exam

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current is used in a heater, electrical energy is


atomic weapons. In the Sun and stars, nuclear fusion reaction occurs in which lighter nuclei like protons, deuterons fuse at high temperature to form a helium nucleus releasing huge amount of energy. At the microscopic level, all different forms of energy are in the form of potential and or kinetic energy.

Equivalence of mass and energy

Albert Einstein gave an equation for inter-conversion of mass and energy as under E = mc2, where c = 3 108 m s is the velocity of light in vacuum.

Conservation of energy

“The total energy of an isolated system remains constant.” This i the statement of law of conservation of energy. One form of energy may get converted into nother form or energy. Energy cannot be created or destroyed. The universe is an isolated system and so the total energy of the universe remains constant.

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omr-conversion of mass and energy as under

“The total energy of an isolated system remains constant.” This i the statement of law of

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om“The total energy of an isolated system remains constant.” This i the statement of law of

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om One form of energy may get converted into nother form or energy.

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omEnergy cannot be created or destroyed. The universe is an isolated system and so the total

www.exam

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omEnergy cannot be created or destroyed. The universe is an isolated system and so the total

physics work

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