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Elodia [21]
3 years ago
6

A 0.70-kg basketball dropped on a hardwood floor rises back up to 65% of its original height. (a) If the basketball is dropped f

rom a height of 1.5 m, how much energy is dissipated in the first bounce? (b) How much energy is dissipated in the fourth bounce? (c) To which type of incoherent energy is the dissipated energy converted?
Physics
1 answer:
dem82 [27]3 years ago
7 0

Answer:

Part a)

Loss = 3.6 J

Part b)

Loss = 0.99 J

Part C)

This is loss in terms of thermal energy due to collision with the floor

Explanation:

Part a)

Since we know that the ball rises up by 65% of initial height

so after first bounce it will lose 35% of its initial energy

so we will have

U = mgH

Energy Loss = 0.35 mgH[/tex]

Loss = 0.35(0.70)(9.81)(1.5)

Loss = 3.6 J

Part b)

Energy of the ball after first bounce

U_1 = 0.65 mgH

energy of ball after 2nd Bounce

U_2 = 0.65(0.65 mgH)

energy of the ball after 3rd bounce

U_3 = (0.65)(0.65^2)mgH

U_3 = 0.65^3(0.70)(9.81)(1.5)

U_3 = 2.83 J

Now we will have energy loss in fourth bounce given as

Loss = 0.35 U_3

Loss = 0.35(2.83)

Loss = 0.99 J

Part C)

This is loss in terms of thermal energy due to collision with the floor

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