Force to hold a spring

**derekjonathon** · December 10th 2009, 08:13 AM

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?

**skeeter** · December 10th 2009, 10:48 AM

Originally Posted by derekjonathon

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?

$<br /> F = kx<br />$

use the 4 lb force and 0.3 ft to determine the spring constant k in lbs/ft.

then integrate the variable force over the interval of the spring's displacement to find the work

$<br /> W = \int_{x_0}^{x_1} kx \, dx<br />$

**novice** · December 10th 2009, 10:57 AM

Originally Posted by derekjonathon

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?

First you must find the spring constant, $k$ , using $F=-kx$ .
You are given $x_1=0.3$ feet, and $F= 4$ lbs.

$k=\frac{-F}{x_1}$ .
The negative sign indicates that the spring is in tension.

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

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

The negative sign indicates work exerted against the sprink energy.

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