AP PHYSICS
posted by Indina on .
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?

Initial energy givenworkAirResistance=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. 
so you're subtracting all of the inital to the final to get the answer?

yes, of course: initialfinal = losses

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