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Chapter 14: Problem 7

During basketball practice Shane made a jump shot, releasing a \(0.60-\mathrm{kg}\) basketball from his hands at a height of \(2.0 \mathrm{m}\) above the floor with a speed of \(7.6 \mathrm{m} / \mathrm{s}\) The ball swooshes through the net at a height of \(3.0 \mathrm{m}\) above the floor and with a speed of \(4.5 \mathrm{m} / \mathrm{s}\). How much energy was dissipated by air drag from the time the ball left Shane's hands until it went through the net?

Short Answer

Expert verified
Answer: Approximately 1.071 Joules of energy were dissipated by air drag from the time the basketball left Shane's hands until it went through the net.

Step by step solution

01

Determine the initial mechanical energy

Calculate the initial mechanical energy (E_initial) of the basketball by considering both its potential energy (PE_initial) and kinetic energy (KE_initial) when released from Shane's hands. Use the following formulas for potential energy and kinetic energy: PE_initial = m * g * h_initial KE_initial = 0.5 * m * v_initial^2 where, m = 0.60 kg (mass of the basketball) g = 9.81 m/s^2 (acceleration due to gravity) h_initial = 2.0 m (initial height) v_initial = 7.6 m/s (initial speed)
02

Calculate the final mechanical energy

Calculate the final mechanical energy (E_final) of the basketball by considering both its potential energy (PE_final) and kinetic energy (KE_final) when it swooshes through the net. Use the following formulas: PE_final = m * g * h_final KE_final = 0.5 * m * v_final^2 where, h_final = 3.0 m (final height) v_final = 4.5 m/s (final speed)
03

Calculate the energy dissipated by air drag

The energy dissipated by air drag (E_dissipated) can be found by subtracting the final mechanical energy of the basketball from its initial mechanical energy: E_dissipated = E_initial - E_final
04

Solve for the amount of energy dissipated

Plug the given values into the equations from Steps 1 and 2, and compute the energy dissipated by air drag using the equation from Step 3: PE_initial = 0.60 kg * 9.81 m/s^2 * 2.0 m = 11.772 J KE_initial = 0.5 * 0.60 kg * (7.6 m/s)^2 = 13.032 J E_initial = PE_initial + KE_initial = 11.772 J + 13.032 J = 24.804 J PE_final = 0.60 kg * 9.81 m/s^2 * 3.0 m = 17.658 J KE_final = 0.5 * 0.60 kg * (4.5 m/s)^2 = 6.075 J E_final = PE_final + KE_final = 17.658 J + 6.075 J = 23.733 J E_dissipated = E_initial - E_final = 24.804 J - 23.733 J = 1.071 J So, the amount of energy dissipated by air drag from the time the ball left Shane's hands until it went through the net is approximately 1.071 Joules.

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Most popular questions from this chapter

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