Elastic Potential Energy - Positive or Negative?

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amandela
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Homework Statement
Q: To stretch a certain nonlinear spring by an amount x requires a force F given by F = 40x - 6x2, where F is in newtons and x is in meters. What is the change in potential energy when the spring is stretched 2 meters from its equilibrium position?
Relevant Equations
F=-kd
INT [-F ]dx = ΔPE
So I understand that I have to integrate the negative of the force function to get the change in PE. I get -(20x^2 - 2x^3) and when I evaluate it from 0 to 2, I get -64N. But, of course, the change is positive. What am I missing?

Thank you.
 
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BvU said:
Make a sketch showing the sign of the directions.Then check your equations.
So the Fs is negative (b/c moving back to 0) and I take the negative integral of the negative function?
 
In the formula,
$$\Delta PE = -\int_{x_0}^x F_{\rm s}(x)\,dx,$$ the force ##F_{\rm s}## is the force exerted by the spring. If you reread the problem statement, the force function ##F(x)## is the force exerted by you (or whatever/whomever is doing the stretching) to stretch the spring.

It's like if you do 10 J of work to lift an object and increase its potential energy by the same amount, the work gravity does is negative because gravity pulls downward but the displacement of the object points upward.
 
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vela said:
In the formula,
$$\Delta PE = -\int_{x_0}^x F_{\rm s}(x)\,dx,$$ the force ##F_{\rm s}## is the force exerted by the spring. If you reread the problem statement, the force function ##F(x)## is the force exerted by you (or whatever/whomever is doing the stretching) to stretch the spring.

It's like if you do 10 J of work to lift an object and increase its potential energy by the same amount, the work gravity does is negative because gravity pulls downward but the displacement of the object points upward.
OK. Thank you.