- #1
Jrlinton
- 134
- 1
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.