# Energy of Motion

So how much energy do we put into an object if we push it for a while? Let's try another simple example, this time in one dimension.

Consider an orange, which we have smeared with a special grease so there is no friction. It is initially at rest, so *v*_{1 }= 0. We push it in one direction with a force *F* over a distance Î”*x*. What is the work done by the force, in terms of the mass *m* and final velocity *v*_{2 }of the orange?

We have an expression for the force given by the second law of motion:

Also we have

Since cos*Ï†* = 1, we have

Notice that the factor Î”*t* drops out.

If we push on an orange initially at rest until it is going at velocity *v*_{2}, then the amount of work we have done on it is 1/2 *mv*_{2}^{2}. This indicates that we can define the *kinetic energy*, the energy of an object due solely to its motion, as

â€¦â€¦(3)

Wow! This is just the kind of expression we saw near the beginning of the Chapter 8. We have completed the circle. But now we have to find out how to use this expression.

Work and change in kinetic energy are related by the following expression.

Work-energy theorem, simple versionIf the total work done on an object is W _{lol}, then its change in kinetic energy is given byâ€¦.(4) |

What is the change in kinetic energy for the woman's cart in the previous section?

We calculated W_{tot} = 0, which tells us that the change in kinetic energy is zero. The cart is going the same speed at the end of the problem as at the beginning. Thus we have Î”*E*_{K} = 0. So this result is consistent with the above equation.

What is the change in kinetic energy of the Earth in one day?

According to the previous section, W_{tot} = 0, and indeed the Earth's speed is constant from one day to the next. The kinetic energy change is zero. Here we are assuming a circular orbit, which is almost correct.

A bullet of mass 20 grams is fired from a gun, so that its speed is 700 m/s in air. The bullet enters a tree stump and embeds 2 meters inside. What is the average force exerted by the stump on the bullet? (Ignore gravity.)

First, we DRAW A DIAGRAM (Figure 9-5). At first this problem looks like a momentum conservation problem, because of the collision and crunching of wood. If we try to apply conservation of momentum, however, we just do not get anywhere.

**Figure 9-5**

The key is to notice that force and distance are both mentioned in the problem.

*Immediately*we think of energy. We know the change in kinetic energy of the bullet

*E*

_{K}=

*E*

_{K2}â€“

*E*

_{K1}

= 0 J â€“ 1/2 (.020 kg)(700 m/s)

^{2}

= â€“4.9 x 10

^{3}J.

This is also

*W*

_{tot}, and we can set

*W*

_{tot }to

*F*Î”

*x*cos

*Ï†*, where cos

*Ï†*is â€“1. Thus â€“4.9 x 10

^{3}J =

*F*(2 m)(â€“ 1),

*F*= 2.45 x 10

^{3}N.

Sometimes, as an object moves along a path, the net force on it changes, or else the displacement changes direction. We can still consider calculating the total work on the object.

**Work-energy theorem, complicated version**

If an object moves along a path, wc can calculate the total work done on the object by dividing the path into tiny pieces and calculating the work done for each piece. The total work W

_{tot}is the sum of the work for these pieces. The change in kinetic energy is given by