A force of 4 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 0.9 feet beyond its natural length?

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- Dec 10th 2009, 08:13 AMderekjonathonForce to hold a spring
A force of 4 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 0.9 feet beyond its natural length?

- Dec 10th 2009, 10:48 AMskeeter
- Dec 10th 2009, 10:57 AMnovice
First you must find the spring constant, $\displaystyle k$, using $\displaystyle F=-kx$.

You are given $\displaystyle x_1=0.3$ feet, and $\displaystyle F= 4$ lbs.

$\displaystyle k=\frac{-F}{x_1}$.

The negative sign indicates that the spring is in tension.

You are also give $\displaystyle x_2=0.9$ feet, which is how much the spring is stretched beyond its natural length.

Work = $\displaystyle \int dW=\int_{0}^{x_2} F dx =-\int_{0}^{x_2} kx dx $

The negative sign indicates work exerted against the sprink energy.