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.