Energy, Power, Work related question

[alevis]

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.
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

 

[tiny-tim]

Welcome to PF!

Hi alevis! Welcome to PF!

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?
…
3.vf = sqrt (1/2mvi^2+mgh-mgh)/m

What about the work done by the pitcher?

(and you're missing a 2)

 

[alevis]

thanks a lot