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Calculating the distance the spring was compressed

  • #1
A 1300 g mass is on a horizontal surface with μk = 0.380, and is in contact with a compressed massless spring with a spring constant of 600 N/m. When the spring is released, it does 8.61 J of work on the mass while returning to its equilibrium position. Calculate the distance the spring was compressed.

Also, what is the velocity of the mass just as it loses contact with the spring?



Any help would be appreciated, I'm pretty lost with this one.
 

Answers and Replies

  • #2
Ackbach
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Follow your problem-solving strategy. What are your target variables? What are the physical principles involved, do you think?
 
  • #3
SammyS
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A 1300 g mass is on a horizontal surface with μk = 0.380, and is in contact with a compressed massless spring with a spring constant of 600 N/m. When the spring is released, it does 8.61 J of work on the mass while returning to its equilibrium position. Calculate the distance the spring was compressed.

Also, what is the velocity of the mass just as it loses contact with the spring?

Any help would be appreciated, I'm pretty lost with this one.
Hello Line6spider. Welcome to PF !

What have you tried?

Where are you stuck?

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  • #4
rude man
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1. all the spring energy was imparted to the mass in one way or another, so how far was the spring compressed?
2. before spring release, all the energy was stored in the spring. After release, part of the energy went into friction loss and part remained in the form of kinetic energy. What was the friction energy loss?

To answer the latter questinn - at what point is contact lost between the spring and the mass?
 
  • #5
Hello Line6spider. Welcome to PF !

What have you tried?

Where are you stuck?
Well someone told me I could use 1/2k x^2= PE but it didn't really work out.

Also, I tried using F=ma and Work is Force times Distance. So F=ma=W/d, but im not sure where to go from there.
 
  • #6
SammyS
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Well someone told me I could use 1/2k x^2= PE but it didn't really work out.

Also, I tried using F=ma and Work is Force times Distance. So F=ma=W/d, but I'm not sure where to go from there.
For a spring it is true that PE = (1/2)kx2 . --- well, of course that is if x is the length by which the spring is compressed (or stretched).

Let's see how you actually went about trying to use that, so we can tell where you're going wrong.
 
  • #7
Would the PE equal the work in this case?

I tried (1/2)kx2=8.61J

Then I could say X=√((2*8.61)/600) right?
 
  • #8
rude man
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Would the PE equal the work in this case?

I tried (1/2)kx2=8.61J

Then I could say X=√((2*8.61)/600) right?
Excellent start!
Now, what happened to that energy?
 
  • #9
SammyS
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Would the PE equal the work in this case?

I tried (1/2)kx2=8.61J

Then I could say X=√((2*8.61)/600) right?
The PE is equal to the work done by the spring. There is another important force here which also does work.
 

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