Max Velocity While Compressing a Spring?

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SUMMARY

The discussion focuses on calculating the maximum compression of a spring when a 17 kg box slides down a frictionless ramp and collides with a spring of spring constant 210 N/m. The maximum compression of the spring was correctly determined to be 2.2 m. However, the participants debated the point of maximum speed of the box, with confusion surrounding the relationship between spring compression and acceleration. The consensus is that the box continues to accelerate until the forces balance, specifically when mgsin(theta) equals kx.

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peterbishop
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Homework Statement



A 17 kg box slides 4.0 m down the frictionless ramp shown in the figure, then collides with a spring whose spring constant is 210 N/m. (picture attached)

a) What is the maximum compression of the spring? (solved this part)
b) At what compression of the spring does the box have its maximum speed?


Homework Equations



F = -kx
Kspring = 1/2 * k* x^2


The Attempt at a Solution



I solved part a already and the answer was 2.2 m, so that's all good. Part b I'm having some trouble with. I've seen this problem attempt to be answered on other threads but no one ever discusses part b thoroughly, it's usually about part a, which I didn't have a problem with. I've seen quite a few people on the internet say that the answer is zero compression, because the spring will slow the box down when it hits it, but this is not the case, as the spring only lessens the acceleration: it is still accelerating, I believe until a point when mgsin(theta) = kx, x being the compression of the spring. Using this I get an answer of .023 m though, which isn't correct either. Any help would be appreciated. (Also I tried putting zero for the answer just for the heck of it but masteringphysics says it's wrong as well.)
 

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peterbishop said:
I've seen quite a few people on the internet say that the answer is zero compression, because the spring will slow the box down when it hits it, but this is not the case, as the spring only lessens the acceleration: it is still accelerating, I believe until a point when mgsin(theta) = kx, x being the compression of the spring.
Sounds good to me.
Using this I get an answer of .023 m though, which isn't correct either.
How did you get that answer? Redo your calculation.
 
Yep you're right, thanks, for some reason whenever I wrote it down in my notes before I did the calculation I left out mass.
 

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