Calculating work done by a spring when stretched

[vinnyzwrx]

Homework Statement

At a certain point, when a spring is stretched near its elastic limit, the spring force satisfies the equation

F = −α x + β x3 , where α = 12 N/m and β = 890 N/m3 .

Calculate the work done by the spring when it is stretched from its equilibrium position to 0.15 m past its equilibrium.
Answer in units of mJ.

Homework Equations

The Attempt at a Solution

since this is a nonlinear curve i thought that the area under the curve would equal the work done by the spring at the .15 m

so i integrated and got

-6x² + 222.5x⁴

since the spring is starting at the equilibrium position, initial position is 0 and final position is .15m so

-6(.15)²+222.5(.15)⁴ = -.02235

this doesn't make sense since we are pulling the spring and the work should be positive right?

 

[PhanthomJay]

Yeah, but the problem asks for the work done by the spring, not the work done by the person pulling it.

 

[vinnyzwrx]

ok so maybe its negative because the spring would be working to go back to its equilibrium position. My answer is wrong though, what am i doing wrong?

am i right in my approach? i really can't think of another way of going about it...

a hint would be very nice

 

[vinnyzwrx]

if work = 1/2kx^2 then all i need to find is the constant k given the information... i don't see how i can do this because i don't know x for the given equation, it just tells me that its near its elastic limit

 

[PhanthomJay]

Work (actually, potential energy) does not equal 1/2 kx^2 in this case. That equation applies only when F = -kx per Hookes law. This spring does not follow that law. Your original solution looks correct to me, except you answered in joules , but the problem asked for the answer in milli-joules (mJ).
-0.02235 J = ____?____ mJ ? (round it off to the nearest whole number).

 

[vinnyzwrx]

-22.35. Thanks a lot, I can't believe I didn't see that.