# Hooke'sLaw

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• May 3rd 2010, 05:26 PM
victorxz90044
Hooke'sLaw
if a force of 12 N stretches a Hooke's law spring .20m beyond its natural length, how much work is done by the agent that does the stretching....
please show steps.
• May 3rd 2010, 05:32 PM
skeeter
Quote:

Originally Posted by victorxz90044
if a force of 12 N stretches a Hooke's law spring .20m beyond its natural length, how much work is done by the agent that does the stretching....
please show steps.

$F = kx$ ... use the given force and displacement to find the spring constant, $k$ , in Newtons/meter.

work done by a variable force ...

$W = \int_0^L kx \, dx$ , where $L$ = length the spring is stretched or compressed beyond its natural length