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

  • #1
physicsissohard
19
1
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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  • #2
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.
 
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Likes Lnewqban
  • #3
sry it was a typo. there is no MHS
 
  • #4
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?
 

1. What is the difference between force and energy?

Force is a push or pull on an object, while energy is the ability to do work or cause a change in an object's motion. Force is a vector quantity, meaning it has both magnitude and direction, while energy is a scalar quantity, meaning it only has magnitude.

2. When should I use force versus energy in a problem?

The choice between using force or energy in a problem depends on the specific situation and what information is given. If the problem involves calculating the motion of an object, it is often more useful to use force equations. If the problem involves changes in an object's energy, it is more appropriate to use energy equations.

3. How does a compressed spring store energy?

When a spring is compressed, work is done on it, causing potential energy to be stored in the spring. This potential energy is in the form of elastic potential energy, which is the energy stored in an object due to its deformation.

4. Can a compressed spring exert a force?

Yes, a compressed spring can exert a force. This force is known as a restoring force, and it is the force that causes the spring to return to its original, uncompressed state. The magnitude of this force is directly proportional to the amount the spring is compressed.

5. How do I calculate the force of a compressed spring?

The force of a compressed spring can be calculated using Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement from its equilibrium position. The equation is F = -kx, where F is the force, k is the spring constant, and x is the displacement from equilibrium.

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