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# Work - kinetic energy theorem

Module by: Sunil Kumar Singh. E-mail the author

Summary: The kinetic energy of a particle changes by the amount of work done on it.

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Work is closely related to energy. Its relationship with energy will automatically come to the fore as we investigate other forms of energy. It is, therefore, imperative that work has similar relationship with kinetic energy. The work - kinetic energy relationship is the subject matter of this module.

To appreciate the connection between work and kinetic energy, let us consider a motion of a block on a rough horizontal plane with a speed "v" in a straight line. The kinetic friction opposes the motion and eventually brings the block to rest after a displacement say "r". Here,

Kinetic friction,

F k = μ k N = μ k m g F k = μ k N = μ k m g

The deceleration, a, is :

a = F k m = μ k m g m = μ k g a = F k m = μ k m g m = μ k g

Now considering motion is x-direction and using equation of motion, v 2 2 = v 1 2 + 2 a r v 2 2 = v 1 2 + 2 a r , we have :

0 = v 2 - 2 a r v 2 = 2 a r = 2 μ k g r 0 = v 2 - 2 a r v 2 = 2 a r = 2 μ k g r

Thus, kinetic energy of the block in the beginning of motion is :

K = 1 2 m v 2 = 1 2 x m 2 μ k g r = μ k m g r K = 1 2 m v 2 = 1 2 x m 2 μ k g r = μ k m g r

A close inspection of the expression of initial kinetic energy as above reveals that the expression is equal to the magnitude of work done by the kinetic friction to bring the block to rest from its initial sate of motion. The work done by the kinetic friction is :

W F = F k r = μ k m g r W F = F k r = μ k m g r

This sums up to a new definition of kinetic energy : Kinetic energy of a particle is equal to the amount of work done by an external force to bring the particle to rest from its state of motion.

## Work - kinetic energy theorem

Work - kinetic energy theorem is a generalized description of motion. Kinetic energy is not only related to work with respect to the action of bringing the particle to rest, but to work, in general, that results in change of the speed of particle. We shall, here, formally write work - kinetic energy theorem for a general description of motion, involving change in kinetic energy of the particle resulting from application of a constant external force. At the end of this module, we shall extend the concept to variable force as well.

Let us first do this for a constant force system with resultant force, "F", as applied on a particle or particle like body. The acceleration of the particle is :

a = F m a = F m

Let the initial and final velocities of the particle be vi and vf. Then using equation of motion, v f 2 = v i 2 + 2 a r v f 2 = v i 2 + 2 a r ,

v f 2 - v i 2 = 2 ( F m ) r v f 2 - v i 2 = 2 ( F m ) r

Multiplying each term by 1/2 m, we have :

1 2 m v f 2 - 1 2 m v i 2 = 1 2 m x 2 ( F m ) r = F r 1 2 m v f 2 - 1 2 m v i 2 = 1 2 m x 2 ( F m ) r = F r

K f - K i = W K f - K i = W

This is the equation, which is known as work - kinetic energy theorem. In words, change in kinetic energy resulting from external force is equal to work done by the force. Equivalently, work done by the force in displacing a particle is equal to change in the kinetic energy of the particle. The above work - kinetic energy equation can be rearranged as :

K f = K i + W K f = K i + W

In this form, work - kinetic energy theorem states that kinetic energy changes by the amount of work done on the particle. We know that work can be either positive or negative. Hence, positive work results increase in the kinetic energy and negative work results decrease in the kinetic energy by the amount of work done on the particle.

### Example 1

Problem : A block of 2 kg is brought up slowly along an incline of length 10 m and height 5 m by applying an external force. At the end of incline, the block is released to slide down to the bottom. If the coefficient of kinetic friction between surfaces is 0.1, find (i) work done by the gravity during round trip (ii) work done by the friction during round trip (iii) work done by the external force and (iii) kinetic energy of the particle at the end of round trip. (consider, g = 10 m / s 2 m / s 2 ).

Solution : This question is structured to bring out finer points about the work done and kinetic energy. Here displacement is along the incline. Thus, we need to consider the components of forces which act along incline plane. Further, we decide the sign of work by determining whether the component of force and displacement are in same direction or opposite to each other.

(i) work done by the gravity during round trip :

The component of gravity along incline is mg sinθ, acting downward. Work done by gravity during up journey is :

W G(up) = F r = ( - m g sin θ ) x L = - m g L sin θ W G(up) = F r = ( - m g sin θ ) x L = - m g L sin θ

Work done by gravity during down journey is :

W G(down) = F r = ( m g sin θ ) x L = m g L sin θ W G(down) = F r = ( m g sin θ ) x L = m g L sin θ

Total work done during round trip is :

W G(roundtrip) = - m g L sin θ + m g L sin θ = 0 W G(roundtrip) = - m g L sin θ + m g L sin θ = 0

(ii) work done by the friction during round trip :

Friction always acts opposite to displacement. Hence, work done by friction in either direction is negative. Now, kinetic friction is :

F k = μ k N = μ k m g cos θ W F(up) = F r = ( - μ k m g cos θ ) x L = - μ k m g L cos θ W F(down) = F r = ( - μ k m g cos θ ) x L = - μ k m g L cos θ W F(roundtrip) = F r = - 2 μ k m g L cos θ W F(roundtrip) = - 2 x 0.1 x 2 x 10 x 10 x ( 1 - 5 2 10 2 ) = - 40 x 0.87 = - 34.8 J F k = μ k N = μ k m g cos θ W F(up) = F r = ( - μ k m g cos θ ) x L = - μ k m g L cos θ W F(down) = F r = ( - μ k m g cos θ ) x L = - μ k m g L cos θ W F(roundtrip) = F r = - 2 μ k m g L cos θ W F(roundtrip) = - 2 x 0.1 x 2 x 10 x 10 x ( 1 - 5 2 10 2 ) = - 40 x 0.87 = - 34.8 J

(iii) work done by external force during round trip :

External force does work only during up motion. The forces on the block constitue a balanced force system as no acceleration is involved. The net external force is equal to the net component force due to gravity and friction on the block.

W E(roundtrip) = ( m g sin θ + μ k m g cos θ ) L = m g L ( sin θ + μ k cos θ ) W E(roundtrip) = 2 x 10 x 10 x ( 5 10 + 0.1 x 0.87 ) = 200 x 0.87 = 117.4 J W E(roundtrip) = ( m g sin θ + μ k m g cos θ ) L = m g L ( sin θ + μ k cos θ ) W E(roundtrip) = 2 x 10 x 10 x ( 5 10 + 0.1 x 0.87 ) = 200 x 0.87 = 117.4 J

(iv) Kinetic energy at the end of round trip :

Initial kinetic energy of the block is zero. The kinetic energy is increased by the work done by the forces acting on the block. Hence, final kinetic energy is :

K f = K i + W = 0 + 0 - 34.8 + 117.4 = 82.6 J K f = K i + W = 0 + 0 - 34.8 + 117.4 = 82.6 J

For understanding purpose and to have an insight, we must understand that net force during up motion is zero. Hence, work done is zero during upward motion. Thus, kinetic energy of the block increases by work done during downward motion only.

## Context of work - kinetic energy theorem

We need to clarify few important aspects of kinetic energy, work and work - kinetic energy theorem in terms of conservative and non-conservative force systems. In reference to a particular type of force called conservative force (gravity, spring force etc.), we define potential energy. We shall study the concept of conservative force in subsequent module.

We should understand that kinetic energy is not limited to conservative force system like potential energy. Also, work - kinetic energy theorem is a general theorem as applicable to all force types and not limited to conservative force. For this reason, we selected examples involving friction (non-conservative force) to emphasize that the concepts of kinetic energy, work and work-kinetic energy theorem are not limited to conservative force as the concept of potential energy.

Second, we derived the mathematical expression of work-kinetic energy theorem in relation to constant force. We restricted to constant force to keep the discussion simple. We must understand that work-kinetic energy theorem is valid to variable force as well.

### Work - kinetic energy theorem and variable force

Here, we set out to establish work - energy theorem for an external force "F", which varies with displacement (r). We consider a particle of mass "m" moving along a straight line under the application of force "F". Let us also consider that force and displacement are in the same direction. The work done in moving the particle by a small linear distance, "dr", is :

W = F . r = F r cos θ W = F r cos 0 0 = F r W = F . r = F r cos θ W = F r cos 0 0 = F r

Now,

F = ma

W = m a r W = m ( v t ) r W = m ( r t ) v = m v v W = m a r W = m ( v t ) r W = m ( r t ) v = m v v

Integrating for limits corresponding to initial and final position of the particle,

W = W W = v 1 v 2 m v v W = W W = v 1 v 2 m v v

W = 1 2 m v r 2 - v i 2 W = 1 2 m v r 2 - v i 2

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