Question

A basketball player makes a jump shot. The 0.599 kg ball is released at a height of 2.18 m above the floor with a speed of 7.05 m/s. The ball goes through the net 3.10 m above the floor at a speed of 4.19 m/s. What is the work done on the ball by air resistance, a nonconservative force?

Answer 1

Answer:

$W_{drag} = 4.223\,J$

Explanation:

The situation can be described by the Principle of Energy Conservation and the Work-Energy Theorem:

$U_{g,A}+K_{A} = U_{g,B} + K_{B} + W_{drag}$

The work done on the ball due to drag is:

$W_{drag} = (U_{g,A}-U_{g,B})+(K_{A}-K_{B})$

$W_{drag} = m\cdot g\cdot (h_{A}-h_{B})+ \frac{1}{2}\cdot m \cdot (v_{A}^{2}-v_{B}^{2})$

$W_{drag} = (0.599\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (2.18\,m-3.10\,m)+\frac{1}{2}\cdot (0.599\,kg)\cdot [(7.05\,\frac{m}{s} )^{2}-(4.19\,\frac{m}{s} )^{2}]$

$W_{drag} = 4.223\,J$

Answer 2

Answer:

W = -4.22 J

Explanation:

Given

m = 0.599 kg

vi = 7.05 m/s

yi = 2.18 m

vf = 4.19 m/s

yf = 3.10 m

We apply the equations of the Principle of Energy Conservation and the Work-Energy Theorem

W = Ef - Ei

W = (Kf + Uf) - (Ki + Ui)

W = (m/2)(vf² - vi²) + mg(yf - yi)

W = (0.599 kg/2)((4.19 m/s)² - (7.05 m/s)²) + (0.599 kg)(9.81m/s²)(3.10 m - 2.18 m)

W = -4.22 J