Spring Energy?

Question

The force exerted by an unusual spring when it's compressed a distance 11 {\rm cm} from equilibrium is given by F = - kx - cx^3, where k = 304N/m and c = 3.5N/m^3.

Find the energy stored in this spring when it's been compressed 11 {\rm cm}.

Isn't the energy stored equal to

U = 1/2 kx^2

I plugged in 304 for k and .11 for x and go 1.839 which was incorrect. I know this problem looks too easy to be true so I want to know what I did wrong.

Thx guys but ur answer did not work...it made sense to integrate it but 1.84 did not work. i even tried negative 1.84 and even tried to plug in values for the original equation f = -kx but still did not work

Answer 1

F = -kx - cx^3
The work done in moving form x to x + dx is
F*dx = -kx*dx - cx^3dx
The work done in moving from 0 to 0.11 is
∫ [0to0.11] = [0to0.11] ∫ {-kx*dx - cx^3dx}
= [0to0.11] {-kx^2 /2 -3 cx^4/ 4)
= - 304 * 0.11^2 /2 -3* 3.5*0.11^4/ 4)
= -1.84 J
1.84 J is stored in it.

Answer 2

This is a good one. For most normal springs where F = -k x the energy stored would be 1/2 k x^2. This spring clearly is not normal for some reason or another.

You need to remember how potential energy and force relate to one another through calculus.

-dU/dx = F

This would suggest that we should integrate force with respect to x to get our stored energy. Lets see if this works for the normal spring.

-dU/dx = - d/dx (1/2 k x^2) = - k x = F(x)

So it works for the normal spring, lets see what our stored energy should be with our unusual restoring force.

F = - k x - c x^3
U = -integral (-k x - c x^3) dx
U = 1/2 k x^2 + 1/4 c x^4

This gives us a much different stored energy than the normal harmonic oscillator.