Conservation of Energy : Elastic Potential Energy

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.

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a) seems ok, but for b) and c), is the spring aligned with the ramp, so the package continues along the ramp, with friction still acting and potential energy still changing as the spring compresses and rebounds?
BrianGEFF719Author Commented:
>>but for b) and c), is the spring aligned with the ramp, so the package continues along the ramp, with friction still acting and potential energy still changing as the >>spring compresses and rebounds?

To give you a visualization of the situation. Think of an elevator shaft with a spring at the bottom, rotated from 90 degrees straight up to n degrees above the horizontal. Thats what it looks like. The spring is parallel with the ramp and the box slides on the ramp....

Yes, i think they want friction to still be acting as it compresses the spring.

then (120 N/m)(compression)²/2 = 2kg*(9.8m/s²)(4m+compression)sin (53.1) - Frictional Force * (4m+compression)

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BrianGEFF719Author Commented:
What numeric answer do you get for B & C)?

BrianGEFF719Author Commented:
Ozo, also, this kind of threw me the problem the tell you the co-efficent for static friction, but I didnt see where I needed it, was that just a 'confusion' technique?

You might want to check to see if the static friction allows the package to start moving at all from the start of the ramp, or after the spring is fully compressed, but otherwise it shouldn't  affect the package in motion.
BrianGEFF719Author Commented:
Thanks, if you have a chance can you perhaps tell me what you get for the numeric solutions to part b & c?

I'm getting 1.06m and 1.32m
BrianGEFF719Author Commented:
>>I'm getting 1.06m and 1.32m

For parts (b) & (c) did you set the origin at the point where the spring is in equilibrium? Because the sum of (b) & (c) should equal (4.0m) if the origin is at the point of equilibrium, is this correct?

BrianGEFF719Author Commented:
Wait, I ment to say the difference of (c) and 4.00m should be how close it gets to its original position?
I thought (c) was how close it gets to its original position
BrianGEFF719Author Commented:
I was interpreting it as how close it gets to 4.00m from the end of the 'relaxed' spring. where x = 0 is the right side of the spring in equilibrium.
I thought "original position" was where the package was released.
If "original position" was the end of the relaxed spring, then
"how close does it get" would seem to suggest that it does not quite reach it, but since I thought it went beyond there, I would have expected "how far does it get"
BrianGEFF719Author Commented:
After redoing the problem I relized I neglected potential energy due to gravity on the way back up....
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