A basketball player makes a jump shot. The 0.664-kg ball is released at a height of 2.27 m above the floor?

Question

with a speed of 7.00 m/s. The ball goes through the net 3.31 m above the floor at a speed of 5.23 m/s. What is the work done on the ball by air resistance, a nonconservative force?

Hey snirifle I am doing my own homework I already have the answers and am just asking to see if there are easier ways to approach the problem. By the way your answer is totally wrong the correct answer is -0.4193492 J. The person who answered below had the correct solution only did not include the negative sign.

Answer 1

Ei - A= Ef (A = work done by air resistance, "i" means initial and "f" means final)

PEi + KEi - A = PEf + KEf

A = PEi + KEi - PEf - KEf

A = mghi + (1/2)mv²i - mghf - (1/2)mv²f

A = m[ghi + (1/2)v²i - ghf - (1/2)v²f]

A = 0.664[(9.80)(2.27) + (1/2)(7.00²) - (9.80)(3.31) - (1/2)(5.23²)]

A = 0.664[22.2 + 24.5 - 32.4 - 13.7]

A = (0.664)(0.6)

A = - 0.4 J (since the work done decreases the total energy, it's negative)

*** 0.4 J of work are done BY the nonconservative force air resistance ON the ball.


*** Believe it or not, the engineer above me did it completely wrong, even though he understood the concept. First, work and energy are NEVER expressed in Netwons (N), but in other units such as joules (J). Also, when calculating the potential energy, the formula is PE = mgh. The engineer forgot the "g," which is equal to 9.80 m/s^2. I am fresh out of AP Physics, and I got a 5 on the AP test. All throughout physics and AP physics, I was drilled with all kinds of problems. Energy/work problems are my specialty. You can even verify my answer with your friends and your teacher, and you'll see that it is correct, and the engineer's answer is very wrong.

*** To the engineer: Sometimes, even professionals can be wrong. I'm sure you understand all the concepts of physics. You obviously have done a LOT of math and physics in the past, since you're an engineer. However, it's probably been a while, and you may have forgotten a few formulas (and I don't think the potential energy formula can be easily derived).

Answer 2

Vx = Vcos(51), Vy=Vsin(51) The parametric equations for projectiles are: X = Xo + (Vx)t and Y = Yo + (Yo)t + 1/2gt^2. X = 14 and Y = 0 because the ball hits the basket at the same height as it left the players hands. t = time. So, 14 = 0 + Vcos(51)t and 0 = 0 + Vsin(51)t + 1/2(-9.8)t^2. You can solve for t in the x-equation : t = 14/Vcos(51) and plug it into the y-equation --> 0 = 14Vsin(51)/Vcos(51) +1/2(-9.8)(14^2/(Vcos(51))^2). The V's cancel in the first part and sin/cos = tan. This simplifies to: 0 = 14tan(51) -68.6/((V^2)(cos^2(51)). By rearranging the equation you can solve for V, which turns out to be V=3.17 m/s.

Answer 3

I would only caution you to be careful with significant digits in the final answer. Since your highest precision was only 3 digits, the solution is only reliable to that number as well, and should be stated as -0.419.