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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.
(a) How much mechanical energy was lost during the collision with the floor?
(b) A basketball player dribbles the ball from a height of 1.37 m by exerting a constant downward force on it for a distance of 0.132 m. In dribbling, the player compensates for the mechanical energy lost during each bounce. If the ball now returns to a height of 1.37 m, what is the magnitude of the force?

Respuesta :

whitneytr12

Answer:

a)[tex]|\Delta E|=4.58\: J[/tex]  

b)[tex]F=61.90\: N[/tex]

Explanation:

a)

We can use conservation of energy between these heights.

[tex]\Delta E=mgh_{2}-mgh_{1}=mg(h_{2}-h_{1})[/tex]  

[tex]\Delta E=0.608*9.81(0.6026-1.37)[/tex]

Therefore, the lost energy is:

[tex]|\Delta E|=4.58\: J[/tex]  

b)

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

[tex]W=\Delta E[/tex]

[tex]F*d=mgh[/tex]

Let's solve it for F.

[tex]F=\frac{mgh}{d}[/tex]

[tex]F=\frac{0.608*9.81*1.37}{0.132}[/tex]

Therefore, the force is:

[tex]F=61.90\: N[/tex]

I hope is helps you!

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