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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 itself energy, but plays a specific role with respect to other forms for energy. Its relationship with different energy forms will automatically come to the fore as we investigate them. In this module, we shall investigate the relationship between work and kinetic energy.

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

Figure 1: Friction applies in opposite direction to displacement .
A block is brought to rest by friction
 A block is brought to rest by friction  (wet1.gif)

Here, kinetic friction is equal to the product of coefficient of kinetic friction and normal force applied by the horizontal surface on the block,

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

Kinetic friction opposes the motion of the block with deceleration, a, :

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 in 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 calculated 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 magnitude of 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 brings up to a new definition of kinetic energy :

Definition 1: Kinetic energy
Kinetic energy of a particle in motion is equal to the amount of work done by an external force to bring the particle to rest.

Work - kinetic energy theorem

Work - kinetic energy theorem is a generalized description of motion. 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, "F", as applied on a particle on a smooth surface. Let v i v i be the initial speed of the particle, when we start observing motion. Now, the acceleration of the particle is :

Figure 2: Force does work on the block.
A force moves the block on a horizontal surface
 A force moves the block on a horizontal surface  (wet2.gif)

a = F m a = F m

Let the final velocity of the particle be v f v f . 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 application of an external force is equal to the 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, compute work by the net force 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 equate with the change in kinetic energy. Most favour this second approach as it does not involve vector consideration with a coordinate system.

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.

We should know that application work-kinetic energy theorem is not limited to cases where initial and final velocities are zero, but can be applied also to situations where these velocities are not zero. We shall discuss these applications with references to specific forces like gravity and spring force in separate modules.

Example 1

Problem : A block of 2 kg is pulled up along a smooth 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. Find (i) work done by the external force and (ii) 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 - kinetic energy" theorem. There are three forces on the block while going up : (i) weight of the block, mg, and (ii) normal force, N, applied by the block and (iii) external force, F. On the other hand, there are only two forces while going down. The force diagram of the forces is shown here for upward motion of the block.

Figure 3: Known forces acting on the block
A block on an incline
 A block on an incline  (wet3.gif)

The most striking aspect about the external force is that we do not know either about its magnitude or about its direction. We have represented the external force on the force diagram with an arbitrary vector. The external force acts only during up journey.

The kinetic energy in the beginning and at the end of "up" motion are zero . Note the wordings of the problem that emphasizes this. From "work - kinetic energy" theorem, we can coclude that sum of the work done by the three forces is equal to zero. It is, now, imperative that if we know the work done by the other two forces, then we shall find out the required work done by the external force, F.

Work done by the forces normal to the incline is zero. Hence, we should only consider force(s) parallel to incline. The component of weight parallel to incline is directed downward. It means that it (gravity) does negative work on the block while going up and positive work while going down.

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) = - m g L sin θ = - 2 x 10 x 10 x ( 5 10 ) = - 100 J W G(up) = F r = ( - m g sin θ ) x L = - m g L sin θ W G(up) = - m g L sin θ = - 2 x 10 x 10 x ( 5 10 ) = - 100 J

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) = m g L sin θ = 2 x 10 x 10 x ( 5 10 ) = 100 J W G(down) = F r = ( m g sin θ ) x L = m g L sin θ W G(down) = m g L sin θ = 2 x 10 x 10 x ( 5 10 ) = 100 J

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

(i) work done by external force ( W F W F ) during round trip :

According to "work-kinetic energy" theorem for motion up the incline :

W G(up) + W F = 0 W F = - W G(up) = - ( - 100 ) = 100 J W G(up) + W F = 0 W F = - W G(up) = - ( - 100 ) = 100 J

(ii) 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 = K f - 0 = W G(round-trip) + W F(round-trip) = 0 + 100 K f = 100 J K f - K i = K f - 0 = W G(round-trip) + W F(round-trip) = 0 + 100 K f = 100 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. Net work is done in downward motion only, whereupon kinetic energy of the block increases.

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.

We should understand that kinetic energy and work are general concept not limited to conservative force. Also, work - kinetic energy theorem is a general theorem as applicable to all force types and not limited to conservative force. For this particular reason, we selected some 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.

Second, we derived the mathematical expression of work-kinetic energy theorem in relation to constant force. We restricted our consideration 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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