CONTINUED:
*i forgot to mention that the ball has a mass 5kg
Basically I'm confused as to how to calculate the potential energy of the ball at point A;
I know that PE = mgh
so PE = (5kg) (9.8 m\s^2) (h) but would h equal the diameter of the circular path? but this doesnt seem right to me.....
hi i am not sure how to do this...thank you
a man is going to pitch a ball. he moves the ball in a circular path of radius 0.6 m, constantly applying a force of 30N to the ball in the same direction as the ball's velocity along the circular path. at the highest point on the path(point A), the velocity of the ball is 15 m\s. what will the speed of the ball be when the man lets go of the ball at the lowest point along the circular path(point B)?
I know that only forces thta are parallel to the motion of an object do work on the object, so in this problem, the man's force of 30N always does work on the ball-- I think this problem has something to do with finding the potential and kinetic energy of the ball at point A and then adding these values (KE + PE) and setting this value for the total energy equal to the KE kinetic energy at point B (since the potenial energy will = 0 at point B)
1 answer
1 answer