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kvv77 [185]
3 years ago
14

A basketball of mass 0.608 kg is dropped from rest from a height of 1.37 m. It rebounds to a height of 0.626 m.

Physics
1 answer:
kenny6666 [7]3 years ago
6 0

Answer:

a)|\Delta E|=4.58\: J  

b)F=61.90\: N

Explanation:

a)

We can use conservation of energy between these heights.

\Delta E=mgh_{2}-mgh_{1}=mg(h_{2}-h_{1})  

\Delta E=0.608*9.81(0.6026-1.37)

Therefore, the lost energy is:

|\Delta E|=4.58\: J  

b)

The force acting along the distance create a work, these work is equal to the potential energy.

W=\Delta E

F*d=mgh

Let's solve it for F.

F=\frac{mgh}{d}

F=\frac{0.608*9.81*1.37}{0.132}

Therefore, the force is:

F=61.90\: N

I hope is helps you!

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