Question

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.

Answer 1

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!