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Nonlinear spring energy problem

  • Thread starter lzh
  • Start date
  • #1
lzh
111
0

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?
 

Answers and Replies

  • #2
F=-dU/dx

use this
 
  • #3
lzh
111
0
So it'd be:
56N(2m)=-dU
dU=-112?
Am I not getting this?
 
  • #4
Astronuc
Staff Emeritus
Science Advisor
18,704
1,718

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) = [tex]\int^x_0\,F(x) dx\,=\,\int^x_0\,kx\,dx[/tex]

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

http://hyperphysics.phy-astr.gsu.edu/hbase/pegrav.html#pe
 
  • #5
lzh
111
0
Oh, I see!

Thanks, I got it!
 
  • #6
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
 

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