Example: Mechanical energy loss

Understanding the situation

In order to see how the work-energy theorem applies when there are resistive forces, let's consider a simple example that lies close to the heart of our everyday experience with motion: If we push a block along a table, it speeds up while we're pushing it, then it slows down and comes to a stop. We've discussed the hidden forces (friction) responsible for this. Let's see how it plays out in terms of energy.

Presenting a sample problem

Consider a system consisting of a metal block. The block has been sitting on a table for a while and is then pushed, sliding on the table and slows down until it comes to a stop.

A. Tell the story of the motion in terms of what forces act on the block, what work is done by which forces, and how those quantities compare.

B. If the magnitude of the force of the hand is constant and equal to $F$, the magnitude force of the friction is constant and equal to $f$, and the hand pushed the block for a distance, $L$, find the distance, $x$, that the block will travel before it stops.

Solving this problem

In analyzing what's happening, let's refer to our anchor equation for this situation: the work-energy theorem.

$$\Delta \big({1 \over 2} \;  m_{B} v_{B}^2 \big) = \overrightarrow{F}^{net}_{B} \cdot  \overrightarrow{\Delta r}_A$$

A, First tell the story.

1. The block starts at rest so its KE is 0.
2. The hand exerts a force on the block and starts pushing it. As the hand moves the block, the force is exerted for a distance and since the direction of the change of position is the same as the direction of the force, the work is positive and tends to increase the  block's KE.
3. There is a friction force on the block that opposes the sliding of the surfaces over each other. Since the hand is pushing the block forward, the friction force is backwards. The friction force is opposite to the change of direction of the block so it is doing negative work on the block and tends to decrease the block's KE.
4. Since the block is accelerating to the right, the force from the hand must be greater than that of friction. Therefore the work done by the hand is greater than the work done by friction so the block increases it's KE, but only by the difference of the two works.
5. The hand stops pushing. Block has some KE, but there is still a force on it: friction. It acts to the left, opposite to the direction that the block is moving, so it does negative work on the block, which eats the block's KE, transferring it to thermal energy until the block comes to a stop.

B. Now lets write the equations that show this result and allow us to calculate $x$.

While the block is accelerating, there are two forces on it doing work: the force of the hand and the force of friction. The work done by each is the force times the distance. So the KE after the pushing phase is

$$KE_1 = (F - f)L$$

After the hand stops pushing, only the friction is acting. The work done by friction in the slowing down phase changes the KE back to 0. Therefore the work done by friction must satisfy

$$ KE_1 = fx$$

So equating these we can find that

$$(F - f)L = fx$$

or

$$x = \frac{(F-f)L}{f}$$

Notice that we did NOT simply plug into the work-energy equation. We focused on the physical content of the terms and decided what works had to be equal. This made handling the signs in the problem (gaining or losing KE) less tricky. 

We can do it directly using the equation by saying that the total work done by the forces must be 0 after the distance $L+x$ since the box started at KE = 0 and ended at KE = 0. The work done by the hand is $W_h = FL$ while the work done by friction is $W_f = -f(L+X)$. Therefore the work-energy theorem gives

$$0 = FL - f(L +x)$$

Solving for $x$ gives the same result. 

It's useful to be able to use both kinds of reasoning and to see how they translate into one another.

Joe Redish 2/24/19