Calculating the distance the spring was compressed

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Homework Help Overview

The discussion revolves around a physics problem involving a mass on a horizontal surface connected to a compressed spring. The mass is subjected to kinetic friction, and the work done by the spring is given. Participants are tasked with calculating the distance the spring was compressed and the velocity of the mass as it loses contact with the spring.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between potential energy stored in the spring and the work done on the mass. Questions arise about the energy distribution between kinetic energy and friction losses. Some participants suggest using energy equations and force relationships to approach the problem.

Discussion Status

There is an ongoing exploration of the problem with various approaches being discussed. Some participants have attempted calculations using potential energy formulas, while others are questioning the assumptions about energy conservation and friction. Guidance has been offered regarding the relationship between potential energy and work done.

Contextual Notes

Participants note the importance of showing prior attempts at solving the problem, as per forum guidelines. There is also mention of the need to clarify the point at which contact between the spring and the mass is lost, which may affect the energy calculations.

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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.
 
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Follow your http://www.mathhelpboards.com/f22/problem-solving-strategy-28/. What are your target variables? What are the physical principles involved, do you think?
 
Line6spider said:
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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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?
 
SammyS said:
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 I am not sure where to go from there.
 
Line6spider said:
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.
 
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?
 
Line6spider said:
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?
 
Line6spider said:
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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