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AP PHYSICS

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A basketball player makes a jump shot. The 0.700 kg ball is released at a height of 1.90 m above the floor with a speed of 7.00 m/s. The ball goes through the net 3.10 m above the floor at a speed of 3.70 m/s. What is the work done on the ball by air resistance, a nonconservative force?

  • AP PHYSICS - ,

    Initial energy given-workAirResistance=finalKE+finalPE

    Think why that is correct.
    Then, fill in the terms.
    InitialKE=1/2 m*7^2
    InitialPE=mg(1.90)
    final KE= 1/2 m 3.7^2
    final PE=mg*3.10
    If you need, I can critique your work.

  • AP PHYSICS - ,

    so you're subtracting all of the inital to the final to get the answer?

  • AP PHYSICS - ,

    yes, of course: initial-final = losses

  • AP PHYSICS - ,

    That is not the correct answer. I did (Initial KE+ Initial PE)- (FinalKE+Final PE)= losses

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