Block sliding down slope against spring

  • Thread starter TG3
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TG3

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1. The problem statement, all variables and given/known data

A block with a mass of 4 kg slides from rest a distance of 2 m down a frictionless inclined plane where it encounters a spring. It compresses the spring a distance 1.4 m before stopping. The inclined plane makes an angle q = 35° with the horizontal. What is the value of the spring constant?

2. Relevant equations
F=-kx
F=MA

3. The attempt at a solution
First, the force on the block = Mass (4) times acceleration. (9.81)
The force of the block when it touches the spring is 39.24.
So, the force the spring exerts is equal and opposite.
39.24 = -kx
X=1.4, so it should be easy:
39.24 = -k 1.4
-k = 28.028
k = -28.028
But this is not right....
 
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It appears that you have everything correct except for one part. The part you have wrong is your force. You are saying the force the block exerts on the spring is mg. That is not correct. Since the block is on a ramp, only a portion of mg is put onto the spring. This portion is mgsin(35). You then set this equal to -kx or kx and solve for k.
 

TG3

66
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So, to clarify, you're saying:
sine 35 x 9.81 x 4 = 22.507
22.507 = -k 1.4
16.07 = -k
-16.07 = k?
Because that answer is still being rejected... (both where k is positive and where it's negative.)
 

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