# calc check: hooke's law

posted by
**COFFEE** on
.

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 :)