A 0.49 kg ladle sliding on a horizontal frictionless surface is attached to one end of a horizontal spring (k = 640 N/m) whose other end is fixed. The ladle has a kinetic energy of 260 J as it passes through its equilibrium position (the point at which the spring force is zero). (a) At what rate is the spring doing work on the ladle as the ladle passes through its equilibrium position? (b) At what rate is the spring doing work on the ladle when the spring is compressed 0.68 m and the ladle is moving away from the equilibrium position?
Genereal Physics I - VitaX, Wednesday, October 13, 2010 at 9:49pm
I worked on this somewhat and I came to the conclusion that in part a since it's asking for the Power as it passes through its equilibrium position, the force is 0. Which makes the Power 0 Watts. For part b though I'm a little more confused as to how to solve it. I know I need to find Velocity and the force but going about it is a bit tricky.
I did the following:
V = (2k/m)^(1/2) but I'm not sure what value to use for the kinetic energy
Fs = -kd = -640*.68 = -435.2 N
If I can get the right velocity I should be able to finish the problem.
Genereal Physics I - VitaX, Wednesday, October 13, 2010 at 11:10pm
Ok, I figured it out. I find the potential energy, and subtract the potential energy from the total energy which was given in the problem. With that kinetic energy now I solve for V using kinetic energy equation. From there I'm able to find the Power with F*V. One of the things that was tricking me up was do I have the displacement negative or not. Apparently I do not, else it's wrong.