Finding the spring constant of a spring in a pulley system

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Homework Help Overview

The problem involves a 3kg object attached to a spring over a frictionless pulley, which is released from rest. The object drops 0.1 meters before stopping, and the goal is to find the spring constant of the spring.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss using force balance and conservation of energy to approach the problem. Some question the assumptions about the object's state when it stops, while others suggest that energy considerations are necessary for the solution.

Discussion Status

There is ongoing exploration of different methods to approach the problem, with some participants asserting that the book's answer is correct while others express confidence in their calculations. Multiple interpretations of the problem's conditions are being discussed.

Contextual Notes

Participants note the need for clarification on whether the object is attached to only one spring and question the implications of the object being at rest when it stops.

Prandals
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Homework Statement


Ok so here is the prompt: a 3kg object is fastened to a light spring over a pulley. The pulley is frictionless and its inertia may be neglected. The object is released from rest when the spring is unstretched. If the object drops 0.1 meters before stopping, find the spring constant of the spring

Homework Equations


Spring force=kx
Gravity force=mg

3. The Attempt at a Solution
My attempt: Since the object is at rest i fgured that the force of the spring must be nullifiying the force of gravity. Therefore, forcespring=kx=weight=mg. plugging in the values and solving for K, i got K=294 Newton/meter. However the book says the correct value is 588 which is double mines. I have re-traced my steps and found that there is absolutely no mistake in getting the value 294. Is the book wrong or am i using the wrong method?
 
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The easiest way to go about it is to use the conservation of energy. The ball falls downward, meaning its potential energy decreases. For energy to be conserved, to what form of energy was it converted to?
 
RedDelicious said:
The easiest way to go about it is to use the conservation of energy. The ball falls downward, meaning its potential energy decreases. For energy to be conserved, to what form of energy was it converted to?
so I just use the formula for Work done by non conservative forces Wnon-conservative=0=changeinPE+changePEElastice+changeinKE right?
 
I need one more information to make sure whether you were right. The object is fastened to only one spring over a pulley or there are something more that you didn't mention about. Because if there aren't anything more then your answer is right.
 
And you have to use energy to solve this. Book is right. 588 is the correct answer
 
Prandals said:
so I just use the formula for Work done by non conservative forces Wnon-conservative=0=changeinPE+changePEElastice+changeinKE right?

Looks good
 
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Prandals said:
My attempt: Since the object is at rest i fgured that the force of the spring must be nullifiying the force of gravity.
It depends what you mean by at rest. The question says it stops, but that doesn't mean all motion has ceased. It only means that its velocity is instantaneously zero.
 
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