A basketball of mass 0.80 kg is dropped from rest from a height of 1.22 m. It re

Question

A basketball of mass 0.80 kg is dropped from rest from a height of 1.22 m. It rebounds to a height of 0.80 m.

(a) How much mechanical energy was lost during the collision with the floor?
__________________J

(b) A basketball player dribbles the ball from a height of 1.22 m by exerting a constant downward force on it for a distance of 0.13 m. In dribbling, the player compensates for the mechanical energy lost during each bounce. If the ball now returns to a height of 1.22 m, what is the magnitude of the force?
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Explanation / Answer

1.

Initial Potential Energy = mgh = 0.8 x 9.8 x 1.22 = 9.5648 Joules
Final Potential Energy = mgh' = 0.8 x 9.8 x 0.8 = 6.272 Joules
Therefore loss in Mechanical Energy = 9.5648 - 6.272
= 3.2928 Joules

2.

Initial Potential energy = mgh = 9.5648 Joules

Final Potential Energy if no force is applied = 6.272 Joules

Now loss in mechanical energy is compensated by player

3.2928 = F x s (Work = force x displacement)

F= 3.2928/0.13

F = 25.33 Newtons

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