A 0.600kg ball is released at a height of 2.00m above the floor with a speed of 7.20m/s. The ball goes through

Question

a basketball net 3.10m above the floor at a speed of 4.20m/s. What is the work done on the ball by air resistance?

The answer is 3.79J but I can't seem to calculate this. I don't know if I'm using the right formulas. Can somebody please explain to me?

Answer 1

Ebefore = Eafter + Eloss

mgho + 1/2mvo^2 = 1/2mvf^2 + mghf + Eloss

11.76 + 15.552 = 5.292 + 18.228 + Eloss

27.312 = 23.52 + Eloss

Eloss = 3.79J

Since the work done by friction is the energy lost in the system, that's your answer.

Answer 2

M = .6 kg

h1 = 2 m

v1 = 7.2 m/s

h2 = 3.1 m

v2 = 4.2 m/s

PE_1 + KE_1 = PE_2 + KE_2 + W_air

M*g*h1 + (1/2)*M*(v1)^2 = M*g*h2 + (1/2)*M*(v2)^2 + W_air

W_air = 3.785 J