# Homework Help: Finding the spring constant of a spring in a pulley system

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1. Mar 27, 2016

### Prandals

1. The problem statement, all variables and given/known data
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

2. Relevant 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?

2. Mar 27, 2016

### RedDelicious

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?

3. Mar 27, 2016

### Prandals

so I just use the formula for Work done by non conservative forces Wnon-conservative=0=changeinPE+changePEElastice+changeinKE right?

4. Mar 27, 2016

### The Vinh

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.

5. Mar 27, 2016

### The Vinh

And you have to use energy to solve this. Book is right. 588 is the correct answer

6. Mar 27, 2016

### RedDelicious

Looks good

7. Mar 28, 2016

### haruspex

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.