Wednesday
July 30, 2014

Homework Help: calc check: hooke's law

Posted by COFFEE on Friday, June 29, 2007 at 11:00pm.

A force of 27N is required to maintain a spring stretched from its natural length of 12cm to a length of 15cm. How much work is done in stretching the spring from 15 to 25cm?

and this is what i did.. please check to see if i did it correctly.. thanks :)

f=27N
12cm to 15cm

f(x) = kx, k=spring constant
15-12cm = 3cm = 0.03cm
f(0.03) = 27N
0.03k = 27
k = 27/0.03 = 900

f(x) = 900x
w = integral of 0.03 to 0.13 for: 900x dx
w = 900*((x^2)/2)] from 0.03 to 0.13
w = 450[((0.13)^2)-((0.03)^2)]
w = 7.2J

Ok on k, 900 N/m

Ok on the next.

thanks :)

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