Reading a Potential Energy Graph

[Jrlinton]

Homework Statement

A conservative force F(x) acts on a 2.0 kg particle that moves along an x axis. The potential energy U(x) associated with F(x) is graphed in the figure. When the particle is at x = 2.0 m, its velocity is –1.2 m/s. (a) What isF(x) at this position, including sign? Between what positions on the (b) left and (c) right does the particle move? (d) What is its particle's speed at x = 7.0 m?

Homework Equations

The Attempt at a Solution

So I started with part d and with the force being conservative I added the potential energy of x=2 (-7J) to its kinetic energy (.5*2kg*(-1.2m/s)^2=1.44J) with its added energy being -5.56 J and set that equal to the potential energy of x=7 (-17J) and its kinetic energy (.5*2kg8v^2) and solved for v to get a speed of 3.38 m/s. That was incorrect.

I then used the same basic technique to calculate the force at x=2. I saw that the acceleration was constant in that interval and used another point (x=3) in that interval to get a velocity value (-2.54m/s) and finding the difference between the two squared velocities ((-2.54)^2-(-1.2)^2)=5.01) and dividing by twice the distance (2) i got the acceleration to be -2.51 as the velocities were decreasing. I then mulitplied the acceleration by the mass of 2kg to get -5.02 N. This was also incorrect.

I am completely unsure as to how to answer b and c either which would seem that these two would be the simplest to figure out.

 

[jbriggs444]

So I started with part d and with the force being conservative I added the potential energy of x=2 (-7J) to its kinetic energy (.5*2kg*(-1.2m/s)^2=1.44J) with its added energy being -5.56 J

A native English speaker would say "total energy" rather than "added energy". The term "added energy" would normally be used to refer to only one part of the total.

As I read the graph, the potential energy at x=2 is PE=-7.5 J (plus or minus .1)

You really should write down some equations rather than just explaining in (too few) words what you did.

and set that equal to the potential energy of x=7 (-17J) and its kinetic energy (.5*2kg8v^2)

As I read the graph, the potential energy at x=7 is PE=-17.25 J (plus or minus .1)

Again, it would be better if you wrote down some equations instead of telling us what you did.

and solved for v to get a speed of 3.38 m/s. That was incorrect.

Would it have been correct if you had read different numbers from the graph?

I then used the same basic technique to calculate the force at x=2. I saw that the acceleration was constant in that interval and used another point (x=3) in that interval to get a velocity value (-2.54m/s) and finding the difference between the two squared velocities ((-2.54)^2-(-1.2)^2)=5.01) and dividing by twice the distance (2) i got the acceleration to be -2.51 as the velocities were decreasing. I then mulitplied the acceleration by the mass of 2kg to get -5.02 N. This was also incorrect.

The graph shows potential energy versus position. What line of reasoning were you using to infer a constant acceleration?

As I understand your approach, you are using the graph to read off potential energy, using potential energy to infer kinetic energy, using kinetic energy to infer velocity, and then using the SUVAT equations to infer acceleration from distance and the difference in squared velocity. There is a much simpler way to proceed.

If you write down some equations, that approach may become obvious. Hint: converting from energy to velocity and back to energy is a waste of two steps.

 

[Jrlinton]

Okay I concluded that the force is equal to the negative of the slope of that portion of the graph on part a. And used the correct values for part d. For b and c the object should only be moving when there is a force present? so when the graph was a slope not equal to zero.

 

[jbriggs444]

Okay I concluded that the force is equal to the negative of the slope of that portion of the graph on part a.

Yes, well done.

For b and c the object should only be moving when there is a force present?

Newton's first law says something about motion when a force is not present. You need a different line of reasoning.