A basketball of mass 0.8 kg is dropped from rest from a height of 1.2 m. It rebo

Question

A basketball of mass 0.8 kg is dropped from rest from a height of 1.2 m. It rebounds to a height of 0.6 m. (a) How much mechanical energy was lost during the collision with the floor? (b) Define what is a non-conservative force, and find the work done by non-conservative forces on the ball. (e) A basketball player dribbles the ball from a height of 1.2 m by exerting a constant down ward force on it for a distance of 0.1 m. In dribbling, the player compensates for the mechanical energy lost during each bounce. If the ball now returns to a height of 1.2 m, what is the magnitude of the force?

Explanation / Answer

part b ) is not a part of mentioned question ...please so ask that question seperately.I am solving remaining 2 parts.

m = 0.8 kg

hi =1.2 m

hf =0.6 m

P.E = mgh

change in P.E = mg [hi - hf ]

change in M.E = mg [hi - hf ]

                     = 0.8 kg * 9.8 [1.2-0.6]

Mechanical energy was lost during the collision with the floor = 4.704 J ---------------->>>>>>>>>>>>>answer)

c ) Force = F = mg [hi - hf ] / s

F = 4.704 J / 0.1

F = 47.04 N ----------------->>>>>>>>>>>>>>>>>magnitude of the force (answer)

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