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

W = -4.22 J

Step-by-step 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

Answer 2

`W_(drag) = 4.223\,J`

Step-by-step 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))+ (1)/(2)\cdot m \cdot (v_(A)^(2)-v_(B)^(2))`

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

`W_(drag) = 4.223\,J`