Nonlinear spring energy problem

[lzh]

Homework Statement

The stretch of a nonlinear spring by an amount x requires a force F given by:
F=40x-6x^2
where F is in Newtons and x is in meters.

What is the change in potential energy U when the spring is stretched 2m from its equilibrium position?

Homework Equations

U=.5kx^2
= .5(kx)x
F=kx

The Attempt at a Solution

F=40(2)-6(2)^2=80-24=56N
U=.5(56N)(2)=56J

Thats what I thought would work, but 56J is no the correct answer. What am I doing wrong?

 

[vipulsilwal]

F=-dU/dx

use this

 

[lzh]

So it'd be:
56N(2m)=-dU
dU=-112?
Am I not getting this?

 

[Astronuc]

Homework Statement

The stretch of a nonlinear spring by an amount x requires a force F given by:
F=40x-6x^2
where F is in Newtons and x is in meters.

What is the change in potential energy U when the spring is stretched 2m from its equilibrium position?

Homework Equations

U=.5kx^2
= .5(kx)x
F=kx

The Attempt at a Solution

F=40(2)-6(2)^2=80-24=56N
U=.5(56N)(2)=56J

Thats what I thought would work, but 56J is no the correct answer. What am I doing wrong?

The trick here is to recognize that F=40x-6x^2 which is not the same as F=kx.

Now U(x) = $$\int^x_0\,F(x) dx\,=\,\int^x_0\,kx\,dx$$

http://hyperphysics.phy-astr.gsu.edu/hbase/pespr.html#pe2

http://hyperphysics.phy-astr.gsu.edu/hbase/pegrav.html#pe

 

[lzh]

Oh, I see!

Thanks, I got it!

 

[vipulsilwal]

f=-dU/dx
integrating both sides
20x^2 -2x^3=-U
put x=2
U=-64J

change would be 64...and work done would be 64 J