# Thread: Work to strecth a srping problem

1. ## Work to strecth a srping problem

The problem is: A force of 1 pounds is required to hold a spring stretched 0.3 feet beyond its natural length. How much work (in foot-pounds) is done in stretching the spring from its natural length to 0.7 feet beyond its natural length?

I'm not sure what the first part really means, would it be the constant force? So then I would use W=F*d?

That's what i'm confused about, I did that and got that the work is 0.3. Then plugged it in to get this intergral:

$\displaystyle \int_{0}^{0.7}(0.3x)dx$

Is that right?

2. Originally Posted by twoteenine
The problem is: A force of 1 pounds is required to hold a spring stretched 0.3 feet beyond its natural length. How much work (in foot-pounds) is done in stretching the spring from its natural length to 0.7 feet beyond its natural length?

I'm not sure what the first part really means, would it be the constant force? So then I would use W=F*d?

That's what i'm confused about, I did that and got that the work is 0.3. Then plugged it in to get this intergral:

$\displaystyle \int_{0}^{0.7}(0.3x)dx$

Is that right?
Hooke's Law ...

$\displaystyle k = \frac{F}{x}$

$\displaystyle k = \frac{1}{.3} = \frac{10}{3}$ lbs/ft

$\displaystyle W = \int_0^{0.7} \frac{10}{3} x \, dx$