Posted by **deb** on Thursday, March 11, 2010 at 9:47am.

A force of 10 pounds stretches a spring 2 inches. Find the work done in stretching this spring 3 inches beyond its natural length

- calculus/physics -
**FredR**, Thursday, March 11, 2010 at 6:49pm
Assuming an ideal spring (linear) and using Hooke's law, the force generated by the spring is:

F = -kx (1)

where x is the distance the spring is stretched from the rest position. So, substituting the info into (1) you get:

-10 = -k(2)

or

k = 5 lb/in

work is the 1st integral of force with respect to x, which would be:

W = (1/2)kx^2

= (1/2)5(3)^2

= 22.5 in-lb

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