# Work done to move spring displacement

1. Oct 15, 2011

### gunster

1. The problem statement, all variables and given/known data
A spring has a relaxed length of 5 cm and a
stiﬀness of 95 N/m. How much work must you
do to change its length from 2 cm to 10 cm?

k=95
Lnull=0.05
delta x = .1-.02 = 0.08

2. Relevant equations

F=-kx
W=Fdcostheta

3. The attempt at a solution

I honestly have tried everything and am beginning to think I am way off the mark and missed something. But what i tried was

W = Fdcostheta where F = -kx

Therefore, since force changes direction after the displacement is past the relaxed spring length, i used:

W = -95 * (0.05-0.02) * (0.05-0.02) cos 0 + -95 * (0.1-0.5) * (0.1-0.5) cos 180

But that was apparently completely wrong. any help please?

2. Oct 15, 2011

### DukeLuke

$F = F d \cos \theta$ is only valid if the force is constant over the distance (not a function of x in this case). Your force is a function of x, so you will have to integrate to get the work. It's possible you can solve the problem with a energy approach if integrals are beyond your course material.

$$W = \int F(x) dx$$

3. Oct 15, 2011

### gunster

EDIT: nvm realized my mistake was suppose to subtract

Thanks a lot for reminding me force is not constant XD

Last edited: Oct 15, 2011
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