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 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.

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

yes, of course: initial-final = losses

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

To calculate the work done on the ball by air resistance, we need to determine the change in mechanical energy. The work done by air resistance is equal to the change in mechanical energy minus the work done by other forces (such as gravity).

The mechanical energy of an object is the sum of its kinetic energy (KE) and potential energy (PE). In this case, the potential energy at the start is the gravitational potential energy, and we can assume the potential energy at the end is negligible since the ball is already close to the ground.

The formula for gravitational potential energy is: PE = m * g * h, where m is the mass of the ball, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height.

So, the initial potential energy is: PE_initial = 0.700 kg * 9.8 m/s² * 1.90 m

The initial kinetic energy is given as: KE_initial = 0.5 * m * v_initial^2, where v_initial is the initial velocity.

The final kinetic energy is given as: KE_final = 0.5 * m * v_final^2, where v_final is the final velocity.

The change in mechanical energy is then: ΔE = KE_final - KE_initial

Finally, the work done by air resistance is: Work = ΔE - PE_initial

Plugging in the given values, we can calculate the work done on the ball by air resistance.