Question

A basketball player makes a jump shot. the 0.600-kg ball is released at a height of 2.09 m above the floor with a speed of 7.52 m/s. the ball goes through the net 3.10 m above the floor at a speed of 4.07 m/s. what is the work done on the ball by air resistance, a nonconservative force?

Answer 1

In order to solve the problem, we need to compare the initial and final energies.
E = mgh + mv^2/2 Initial E = m (g* 2.09 +7.52^2/2) Final E = m(g*3.1+4.07 ^2/2) E lost = Initial E- Final E= (1/2(0.600)(4.07)^2 + (0.600)(9.8)(3.10)) - (1/2(0.600)(7.52)^2 + (0.600)(9.8)(2.09)) 
Work done by air on ball is -6.05685 J.