A 0.18 kg ladle sliding on a horizontal frictionless surface is attached to one end of a horizontal spring (k = 680 N/m) whose other end is fixed. The ladle has a kinetic energy of 9.8 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 0.019 m and the ladle is moving away from the equilibrium position?
The Attempt at a Solution
(a) I found that answer is zero. This is because spring has no work done at the particular moment.
(b) I let 1/2mv^2 = 9.8J
Solve for v and I get 10.43m/s.
According to the question (b), it states that need to find the instantaneous power of the spring. Therefore, I let Fs = -Kd
Fs= force from spring
Then, substitute Fs into power formula, which is P= fv . The answer I obtained, is 134.76W.
But the answer is wrong. May I know the reason ? Could it be I mess up with the positive and negative sign ? Isn't "d" negative in this situation because it's being compressed ? Thank you.