This isn't the actual problem given in the assignment. I've twisted it a bit because I want to see were I specifically go wrong in my calculations.(adsbygoogle = window.adsbygoogle || []).push({});

1. A pitcher rotates a 0.25kg ball around a vertcal circular path of radius 0.6m before releasing it. The pitcher exerts a 50N force directed parallel to the motion of the ball around the complete circular path. The speed of the ball at the top of the circle is 20m/s. If the ball is released at the bottom of the circle, what is the speed upon release?

2. KE = 1/2mv^2

PE = mgh

Energy at top = Energy at bottom.

1/2mvi^2+mgh = 1/2mvf^2+mgh.

3.vf = sqrt (1/2mvi^2+mgh-mgh)/m

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# Homework Help: Energy, Power, Work related question

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