A ball of mass M = 0.320kg is connected by a light but rigid rod of length L = 0.770m to a pivot and held in place with the rod being vertical. A wind exerts a constant
force F to the right on the ball. The ball is released from rest, and the wind makes it swing up to a maximum height Hmax above its starting point before it swings down again. (a) Find Hmax when F = 10.0N; (b) Find the equilibrium height, H for F = 10.0N.
1/2 mv^2 = mgh
F = ma
The Attempt at a Solution
Firstly, I found a = 31.3 m/s^2. Then I would like to apply the equation, 1/2 mv^2 = mgh, to solve the question, but I realized that v was still unknown. I have tried several ways to find v. Yet, I failed every time. Can anyone advise me?
Since part a is not solved, I am not able to solve part b. Besides, would anyone mind telling me what is meant by equilibrium height? What are the differences between the equilibrium height and the maximun height?