Calculating Work Done on a Spring with Given Stiffness and Length Changes

[karobins]

Homework Statement

A spring has a relaxed length of 7 cm and a stiffness of 150 N/m. How much work must you do to change its length from 12 cm to 15 cm?

Homework Equations

F=-kx, W=F(delta)x

The Attempt at a Solution

?? ((.12 m)-(.07 m) * (150 N/m)) = 7.5 N

((.15 m)-(.07 m) * (150 N/m)) = 12 N

(12-7.5 N) * (.15-.12 m) = .135 Nm

 

[LowlyPion]

Welcome to PF.

You need to consider that W = ΔPE

What is the formula for the PE of a spring?

 

[karobins]

thanks much for the help!

 

[fball558]

i assume you got the answer but if not
a quick way to solve this problem is to
set up a integral.
integrate Kx where K is the stiffness
so it gives you 75x^2
act like relaxed length is initial (or 0) then find difference
from initial your two asking points are (make sure your in m not cm)
ex. rest is 7cm find work 13cm stretched to 15 cm
first limit would be (.13-.07)=.06 next (.15-.07)=.08
just plug these into 75x^2
do F(b)-F(a) and that gives you the answer