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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.

### Work - kinetic energy theorem with multiple forces

Extension of work - kinetic energy theorem to multiple forces is simple. We can either determine net force of all external forces acting on the particle and then apply work - kinetic energy theorem. Otherwise, we can determine work done by individual forces for the displacement involved and then sum them to find the work done by the multiple forces on the particle.

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

### Application of Work - kinetic energy theorem

Work - kinetic energy theorem is not an alternative to other techniques available for analysing motion. What we want to mean here is that it provides a specific technique to anlayze situations where force is varying or where information on intermediate motion is not available. Here, we shall work with an example where a block is raised up along an incline. We do not have information about the nature of motion - whether it was raised up along the incline slowly or with constant speed or with varying speed. We also do not know - whether the applied external force was constant or varying. But, we know the end conditions that the block was stationary at the beginning of the motion and at the end of motion. So,

K f - K i = 0 K f - K i = 0

It means that work done by the forces on the block should sum up to zero. If we know other forces and hence work done by them, we are in position to know the work done by the "unknown" force and knowing the displacement, we can determine the average force applied on the block during its motion along the incline.

### Example 1

Problem : A block of 2 kg is brought up 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. As discussed earlier, work done by all forces on the block equals to zero. Now, we see that work done by both gravity and friction act downward and as such work done by them during up motion is negative. The work done by the external force, therefore, should be equal to the work done by other forces and opposite in sign so that total work done is zero. It is important to realize that we could not have calculated work by an unknown force without work - energy theorem.

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, we again emphasize that work done during up motion is zero as block is stationary in the beginning and at the end during motion up the incline. Work is done in downward motion only, whereupon kinetic energy of the block changes.

## 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. The concept of potential energy is specific to conservative force.

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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