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?

Answer 1

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.

Answer 2

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

Answer 3

yes, of course: initial-final = losses

Answer 4

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