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?

Question

Benjamin Hood

Physics

24 July, 02:51

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?

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Mcdonald · 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.