Problem with when to use Force and Energy. What compresses spring/

AI Thread Summary
The discussion centers on the confusion regarding the application of force and energy principles in a spring compression problem. The original method used by the participant involved equating the gravitational force to spring force, but their teacher suggested using conservation of mechanical energy instead. The participant understands the energy conservation method but struggles to grasp why their initial approach is incorrect, particularly regarding the relationship between force and compression. Key points include the realization that the force exerted by the spring changes as it compresses, which affects the overall compression distance. Ultimately, the discussion highlights the importance of understanding how varying forces influence spring behavior beyond just Hooke's Law.
physicsissohard
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Homework Statement
A mass of 20 kg is released from rest from the top of a fixed inclined plane of inclination 53 deg and height 4 m. At the bottom of the inclined plane, there is a massless spring of length 2 m. Find the maximum compression of the spring.(k=10000 N/m) take g=10m/s2
Relevant Equations
ME1=ME2
This is how I tried to do it, which is the most direct. The force that the mass exerts on the spring is mgsin(53). and I equated that to kx. and found x. but apparently, this is wrong and the teacher told me a different method.
(ME)1=(ME)2 due to conservation of mechanical energy
20∗10∗4=20∗10(2−x)∗0.8+0+0.5∗10000∗x^2 on LHS there is no kinetic energy and the potential energy of the block is the only thing on the left. And RHS there is no kinetic energy but the potential energy of the block and spring is there. And from here you just need to solve the quadratic. I understand this method and see nothing wrong with but I don't understand what is wrong with mine. I think I even get where the difference is coming actually. It's even more intuitive that in the second method, you observe that when the block collides with the spring it compresses very much and comes back to less than the original length, which doesn't take it into account. But I don't understand why it compresses more cuz the same force is applied, so it moves the same distance. Just can somebody elucidate why force doesn't only determine the compression? It's hookes Law I don't see what's wrong though.
 
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physicsissohard said:
Homework Statement: A mass of 20 kg is released from rest from the top of a fixed inclined plane of inclination 53 deg and height 4 m. At the bottom of the inclined plane, there is a massless spring of length 2 m. Find the maximum compression of the spring.(k=10000 N/m) take g=10m/s2
Relevant Equations: ME1=ME2

20∗10∗4=20∗10(2−x)∗0.8+0+0.5∗10000∗x220∗10∗4=20∗10(2−x)∗0.8+0+0.5∗10000∗x^2 on LHS there is no kinetic energy and the potential energy of the block is the only thing on the left.
Can you rewrite this equation using symbols first and then substitute numbers? I have a hard time deciphering expressions like 0.5∗10000∗x220∗10∗4. Also, I can see a LHS (left hand side) and a RHS (right hand side) but I also see a MHS (middle hand side). What's that about?

Please use the LaTeX editor for your equations.
 
sry it was a typo. there is no MHS
 
physicsissohard said:
But I don't understand why it compresses more cuz the same force is applied, so it moves the same distance. Just can somebody elucidate why force doesn't only determine the compression? It's hookes Law I don't see what's wrong though.
Consider how the force with which the spring reacts changes during the distance x.
Is that force greater or lesser than kx at the beginning of the compression stroke?
 
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