Skater Ramp Simulation: Potential Energy vs. Position

[Ibraheem]

Hello,

In this simulation (link is below) the graph of the potential energy with respect to position U(x) when there is no friction on the ramp looks like the graph of potential energy of Hooke's Law spring U(x)=(1/2)KX^2). For the spring, the negative of the derivative is equal to the horizontal restoring force. So what restoring force is the negative derivative of potential energy graph for the skater in the simulation?I have included a picture of the simulation along with the potential energy U(x) graph.

Simulation:
http://phet.colorado.edu/en/simulation/energy-skate-park

 

[LunaFly]

Potential energy due to gravity is being plotted in this simulation. That being said, it is a tad bit complicated since the skater is on a ramp.

For example, an object in free-fall would have potential energy mgy. The force on the object would then be -d/dy (mgy) = -mg. Simple enough. Notice, however, that the only force acting on this object is that due to gravity.

For the ramp system, (assuming "frictionless" ramp used) the skater being acted upon by both the gravitational force and the normal force of the ramp at all times. This means that the net force on the skater is a sum of these two forces, and so the plotted potential energy function is related to both of these forces.