# Example: Getting forces from PE - 1D

#### Prerequisites

## Understanding the situation

The figure at the right shows a mass attached to a light ("massless") spring whose other end is attached to the wall. The position $x = 0$ corresponds to the relaxed (unstretched) position of the spring. A graph of the PE function for the system is shown below.

## Presenting a sample problem

*Find the directions and relative magnitudes of the forces exerted on the cart when it is at the positions marked by the numbers 2, 1, 0, -1, and -2 from reading the information from the graph.*

## Solving this problem

In 1D this is straightforward. We can use the analogy of a "ball rolling on a hill. The direction of the force is always "down the hill". The equation that codes this in math is

$$F^{spring}_x = -\frac{dU^{spring}}{dx}$$

where in this case, "type" = has become spring (even though we have used a gravitational analogy).

On the positive side, the PE is growing at an increasing rate, so the slope, $dU/dx$, is positive. This means that the force is negative -- pointing to the right (downhill), pulling the mass back towards the origin. And it grows as the mass moves farther out, stretching the spring more.

When the object is at the bottom of the curve, where the PE curve is flat (at $x = 0$) the slope is 0 so there is no force. (A ball could rest there without being pushed to one side or another.) Physically, the mass is at the point where the spring is not stretched, so there is no force.

As the object moves to the left, the slope of the curve goes increasingly negative (i.e., the tangent line points down to the right). So the force is positive -- points to the right (downhill), again, pulling the mass back towards the origin. Again, the force grows in magnitude as you go further from the origin, since as the mass moves farther in, it compresses the spring more.

So the answers are:

- At 1 and 2 the force is negative, pushing the object back toward the origin (to the left).
- At 0 the force is 0.
- At -1 and -2 the force is positive, pushing the object back toward the origin (to the right).
- The magnitudes rank as follows:

$$F_2 = F_{-2} > F_1 = F_{-1} > F_0 = 0$$

Since we actually know the function $U$ explicitly, (it is $U = {1 \over 2}kx^2$) we can find the results by calculation as well, taking the derivative:

$$F = - \frac{d}{dx}({1 \over 2}kx^2) = -{1 \over 2}k\frac{dx^2}{dx}= -{1 \over 2}k(2x) = -kx$$

Check that using this gives the same result as reading the graph.

Joe Redish 11/4/14

#### Follow-on

Last Modified: November 18, 2019