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# Conservation of Energy : Elastic Potential Energy

Posted on 2006-03-26

A 2.00kg package is released on a 53.1 degree inclide 4.00m from a long spring which is attached at the bottom of the incline (so to picture this there is a ramp with a spring and 4m above the end of the spring there sits this package), the force constant of the spring is 120 N/m. The coefficents of friction are us = 0.40 and uk = 0.20, the mass of the spring is negligable.

a) what is the speed of the package before it hits the spring?

h = 4.00m sin (53.1)

Sum Forces in Y Direction = N - mg cos 53.1 = 0; N = mg cos 53.1

Frictional Force = .20(mg cos 53.1)

W = Frictional Force * 4m = 9.41 J

K1 + U1 + W = K2 + U2

U1 + W = K2 + U2

mgh + w = 1/2m(v2)^2

2gh + 2w/m = (v2)^2

v2 = 7.3m/s

Not sure about that answer.

b) what is the maximum compression of the spring?

1/k(m(v1)^2 + 2w = (x2)^2

x2 = -1.02 m

Again not sure.

c) when the package rebounds back up the ramp how close does it get to its original position.

U1 + W = U2

1/2k(x1)^2 + w = 1/2k(x2)^2

(x2)^2 = x1^2 - 2w/k

x2 = .94m, so its 3.06m away from its starting point.

I am not sure about any of these answers, if incorrect please advise what i'm doign wrong.

Brian