What is the maximum compression of the spring

  • Thread starter ja2ha
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  • #1
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Homework Statement


A mass is initially held in place at the top of a spring. When it is let go, it falls and compresses a spring. What is the maximum compression of the spring.


Homework Equations



f=-kx

The Attempt at a Solution


I know the solution is to equate mg and kx to find x. However, it doesn't make sense when I try it using energy equations. If you set 1/2kx^2 to mgh (where h = x), then the answer you get for x is 2 times the answer you get doing it the proper way. Does anybody know why?
 

Answers and Replies

  • #2
LowlyPion
Homework Helper
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Welcome to PF.

I think you are confusing Force with Work.

The Work to compress the spring is ½kx² and the PE from x=0 is as you expect mgx.

But work is the ∫F⋅x dx and the F is a function of x and is equal to -kx.

When you evaluate that you get Wx = ½kx² = PE = m*g*x

If you wanted to properly use the ½kx² then take the derivative at the point x which happily is kx. The PE as a derivative of x is simply m*g.

And look at that x = mg/k
 
  • #3
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Hi thanks for the reply.
You said that

Wx = ½kx² = PE = m*g*x

Why can't you just solve for x in this step? That's what I've been wondering all along?
 
  • #4
LowlyPion
Homework Helper
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Because the force varies along the path as a function of x. And work is the integral of the dot product of F and X. And what you are looking at in the ½kx² is the average force across the distance. (1/2*kx)*x = ½kx²

If you look at the curve, the force is a straight line slope of k in x. And the area is 1/2 base*height which represents the area of the "Average Force" across the distance x and that is what equates to the change in potential energy.

If the Force had been k*h across the whole range of x, then that would equal m*g*h.
 
  • #5
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Because the force varies along the path as a function of x. And work is the integral of the dot product of F and X. And what you are looking at in the ½kx² is the average force across the distance. (1/2*kx)*x = ½kx²

If you look at the curve, the force is a straight line slope of k in x. And the area is 1/2 base*height which represents the area of the "Average Force" across the distance x and that is what equates to the change in potential energy.

If the Force had been k*h across the whole range of x, then that would equal m*g*h.
LP, the problem setup doesn't specify the initial condition of the spring. Assuming the spring is neither compressed nor extended, the equilibrium position of the mass plus sping is -mg/k. The displacement at maxium compression is an additional -mg/k.
 

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