Mass on incline plane sliding into spring, need to find max compression!


by BryceHarper
Tags: compression, max, physics
BryceHarper
BryceHarper is offline
#1
Jun7-12, 08:49 PM
P: 15
1. The problem statement, all variables and given/known data

A 11 kg box slides 4.0 m down the frictionless ramp shown in the figure , then collides with a spring whose spring constant is 190 N/m. The angle of the ramp is 30. What is the maximum compression of the spring?


2. Relevant equations

Ei=mgh
Ef=1/2kx^2

3. The attempt at a solution

Ei=Ef
mgh=1/2kx^2
mgLsinθ=1/2kx^2
11(9.8)(4sin30)=1/2(190)x^2
1.5m = Max compression

I swear this is right but it's not, can anyone help me out?
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Villyer
Villyer is offline
#2
Jun7-12, 08:58 PM
P: 294
Is the spring laying on the ramp, or is there a flat part before the mass hits the spring, making the spring horizontal?

Also, does the 4m refer to the distance it slid along the ramp, or the vertical displacement?
BryceHarper
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#3
Jun7-12, 09:03 PM
P: 15
The spring is laying on the ramp and the 4m refers to the distance it slid along the ramp.

Villyer
Villyer is offline
#4
Jun7-12, 09:08 PM
P: 294

Mass on incline plane sliding into spring, need to find max compression!


You made the statement that the energy of the block 4 meters above the uncompressed spring is equal to the energy of the compressed spring, but that is leaving out a portion of energy.

From the point when the block is released to the time is stopped (at the maximum compression point of the spring) what is its change in height?
BryceHarper
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#5
Jun7-12, 09:16 PM
P: 15
Ohhhhhhhhh!!!

So i should do:
1/2(4+Xmax)sinθ=1/2k(Xmax)^2

?
Villyer
Villyer is offline
#6
Jun7-12, 09:17 PM
P: 294
The left side should read 11(9.8) instead of (1/2), but besides that yes.
BryceHarper
BryceHarper is offline
#7
Jun7-12, 09:27 PM
P: 15
Whoops! I guess I was so excited to have finally figured it out that I wrote the equation wrong lol.

To verify

mg(4+Xmax)sinθ=1/2k(Xmax)^2

So...

1/2k(Xmax)^2 - mgsinθ(Xmax) - mg4sinθ = 0

And then use quadratic equation to solve for Xmax?
Villyer
Villyer is offline
#8
Jun7-12, 09:29 PM
P: 294
That looks right to me.

I wonder what the nature of the roots will be of that equation?
BryceHarper
BryceHarper is offline
#9
Jun7-12, 09:34 PM
P: 15
Got it! Xmax is equal to 1.8m

Thanks for the help Villyer!
Villyer
Villyer is offline
#10
Jun7-12, 09:37 PM
P: 294
Of course


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