A basketball player makes a jump shot. The 0.700-kg ball is released at a height of 2.00 m above the floor with a speed of 7.70 m/s. The ball goes through the net 3.00 m above the floor at a speed of 3.10 m/s. What is the work done on the ball by air resistance, a nonconservative force?

Answer 1

Calculate the loss of total mechanical energy (potential + kinetic) from release until it reaches the basket.

(1/2)M*V1^2 + M*g*H1 = (1/2)MV2^2 + M*g*H2 + (E lost)

Elost = 0.700{[(1/2)(7.7)^2-(3.1)^2] + (9.8)(2.00 - 3.00)} = ____

I consider it unlikely that a basketball would slow down that much due to air resistance, but those are the numbers you were given.

Answer 2

10.5 J = frictional work lost
Initial KE = 20.7 J

Something is phoney here.