# Spring Problem and average force

1. Dec 3, 2007

### temaire

1. The problem statement, all variables and given/known data
http://img88.imageshack.us/img88/9351/springny2.jpg [Broken]​
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2. Relevant equations
$$E_p = \frac{Fx}{2}$$
$$E_k = \frac{mv^2}{2}$$

3. The attempt at a solution
http://img231.imageshack.us/img231/4773/worknd6.jpg [Broken]​
[/URL]

I get 2.22 m/s, but the answer key says 3.1 m/s. Where did I go wrong?

Last edited by a moderator: May 3, 2017
2. Dec 3, 2007

### PhanthomJay

The average force is already given as 5.7N. You mistakenly divided it again by 2. E_p = F_avg(x)

3. Dec 3, 2007

### Galileo's Ghost

Check your value for Force. The problem gives average force...this is not equal to the actual force when the spring is compressed 1.3 cm.

4. Dec 4, 2007

### temaire

So Phanthom, you're saying that $$E_p = F_{ave}x$$ is an actual formula?

5. Dec 4, 2007

### PhanthomJay

The elastic potential energy stored in a spring is equal in magnitude to the work done by it when moving from its compressed distance, x, to its original uncompresed length, that is, $$E_p = \int Fdx$$, from the definition of work, where F, from Hookes Law, is $$F =kx$$. Performing the calculus, you get the well known equation for the elastic potential energy of the spring, $$E_p =1/2kx^2$$. Noting that $$F=kx$$ at its max compressed point, this is identically equivalent to $$E_p = F/2 (x)$$, or $$E_p = F_{avg} x$$.
In your problem, the average force is given as 5.7N. This means the force in the spring is 0 in its uncompressed length, and 11.4N when it is compressed 1.3cm.

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