Example: Potential Energy Graphs

Question:

Suppose an object of mass 10 kg is acted upon by this potential. It is located at x = 6 and moving to the left at v = 0.63 m/s. Where does it turn around and head back in the positive x-direction?

Answer:

Let's do it graphically first. Note that K = 1/2 (10)0.63² = 2.0 J. So we need to find the total energy.

E = U + K

It looks like U(7) = 1.0 J, so the total energy E = 3.0 J. The graph now can show the kinetic energy K in yellow and the total energy E in green.

The green line is the total energy E = 3.0 J. The turning points are by inspection x = 2 and x = 8.

To do the problem more analytically, find the equation which represents the potential energy curve. You only need the left part since you are asked for the first turning point. There are points at (1,4) and (5,0) so you have to solve

4 = a + b(1)
0 = a + b(5)

solving for a and b we get

U(x) = 5 - x

Now you need to find when E = U = 3.0 J

3.0 = 5 - x

x = 2 as above.

Question:

What is the maximum speed during the motion?

Answer:

It occurs when the potential energy is minimum, x = 5. At this point U = 0, so the total energy is all kinetic. K = 3.0 J = 1/2 mv² = 1/2 (10) v². Solving for v = 0.77.