# Force to hold a spring

• Dec 10th 2009, 09:13 AM
derekjonathon
Force 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, 11:48 AM
skeeter
Quote:

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?

$
F = kx
$

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

$
W = \int_{x_0}^{x_1} kx \, dx
$
• Dec 10th 2009, 11:57 AM
novice
Quote:

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.

$k=\frac{-F}{x_1}$.
You are also give $x_2=0.9$ feet, which is how much the spring is stretched beyond its natural length.
Work = $\int dW=\int_{0}^{x_2} F dx =-\int_{0}^{x_2} kx dx$