How Does Potential Energy Influence Force and Motion in Physics Experiments?

[susie__]

Homework Statement

2002M3
An object of mass 0.5 kg experiences a force that is associated with the potential energy function
U(x) = 4/ 2+x , where U is in joules and x is in meters.
a. On the axes below, sketch the graph of U(x) versus x.

(graph) Y aixs U(J), x-axis x(m)b. Determine the force associated with the potential energy function given above.
c. Suppose that the object is released from rest at the origin. Determine the speed of the particle at x = 2 m.

In the laboratory, you are given a glider of mass 0.50 kg on an air track. The glider is acted on by the force determined in part b. Your goal is to determine experimentally the validity of your theoretical calculation in part c.

d. From the list below, select the additional equipment you will need from the laboratory to do your experiment by checking the line to the left of each item. If you need more than one of an item, place the number you need on the line.

Meterstick _____ Stopwatch Photogate timer String Spring

Balance Wood block Set of objects of different masses

Homework Equations

PEg=mgh
.5mv^2=KEe

The Attempt at a Solution

On a I think you just draw the graph so I'm okay with that one. ;)

On b what force are they talking about? I didn't think PEg was a force..

 

[aurora14421]

To answer part (b), the force associated with a potential energy is given by:

F= - dU/dx

So I think you just need to differentiate the equation you're given for the potential energy, U(x), and then take the negative of it.

 

[ogb p]

In this case, the force associated with the potential energy function is the force exerted on the object by the potential energy. This can be calculated using the formula F = -dU/dx, where U is the potential energy function and x is the position of the object. In this case, the force would be F = -(4/2+x)^2 = -(8+4x)^2 N.

For c, we can use the conservation of energy to determine the speed of the object at x = 2 m. Since the object is initially at rest, all of its initial energy is in the form of potential energy, which is given by U(x=0) = 4 J. At x = 2 m, the potential energy is U(x=2) = 4/2+2 = 2 J. Therefore, the kinetic energy at x = 2 m is KE = 4 J - 2 J = 2 J. Using the formula KE = 1/2mv^2, we can solve for the velocity v = √(2/0.5) = √4 = 2 m/s.

For d, the additional equipment needed would be a meterstick (to measure the distance traveled by the object), a stopwatch (to measure the time taken for the object to travel a certain distance), and a set of objects of different masses (to test the validity of the theoretical calculation).