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A basketball player makes a jump shot. The 0.585 kg ball is released at a height of 2.14 m above the floor with a speed of 7.15 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?

  • physics -

    consider U = 0 at 2.14 m height
    total energy at bottom = Ke = (1/2).585(7.15)^2 = 15 J

    height gained = 3.1 - 2.14 = .96 m
    total energy at top
    = .585 g (.96)+.5(.585)(4.19)^2
    = 10.6 J

    difference = energy lost to friction = 15-10.6 = 4.4 J

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