Physics problem, basketball dunk (Hookes law)

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Homework Statement


A 95kg basketball player slam dunks a basketball and hangs onto the rim. Find out how much the rim bends if its spring constant = 7400 N/m and the basketball rim is 2 m in the air.

Homework Equations


Ep = 0.5 k x^2
Ek = 0.5 m v^2
Eg = mgh

The Attempt at a Solution


The book got 0.15 m. My answer is way off, what did I do wrong? I think this should be right.

Ep1 + Ek1 + Eg1 = Ep2 + Ek2 + Eg2
Ep1 = 0, Ek1 = 0, Ek2 = 0

Eg1 = Ep2 + Eg2
mgh = (1/2)kx^2 + mg(2m - x)

1862 J = 3700x^2 N/m + 1862 J - 931x N

3700x^2 N/m - 931x N = 0

931x N( 3.97x / m - 1 ) = 0

x = 0 or 3.97x/m = 1

x = 1/(3.97/m) = 0.25 m.

Book got 0.15. My answer is way off and I'm not sure why it could be look this. Is my logic correct? Am I correct?
 
Last edited:
on Phys.org
You are not given any speeds so there is no point in looking at kinetic energy. The problem is just "if a 95 kg object is hung from a spring with spring constant 7400 N/m, how far will the spring extend?"

95g=