springs

[jeca86]

a spring has a natural length of 20cm. If a 25-N force is required to keep it stretched to a length of 30cm, how much work is required to stretch it from 20cm to 25cm?

ok so i know i have to integrate from x=20 to 25.

So far this is what i have:

25-N=f(x)=kx=.98424k [30cm=.98424ft]
25=.98424k
25.4=k
f(x)=25.4x

Is this right? If so now do I integrate? what are the units for N in 25-N?

 

[skeeter]

why are you changing to English units? do the work in metric ... much easier.

Hooke's Law ...
F = kx

25 N stretches the spring 10 cm = 0.1 m

25 N = k(.1 m)

k = 250 N/m

W = INT{.2 to .25} kx dx

W = [(1/2)kx^2] evaluated from .2 to .25 m by the FTC ...

W = 125[.0625 - .04] = 2.8125 Joules