Calculating work done by a spring when stretched

AI Thread Summary
The discussion revolves around calculating the work done by a spring described by a nonlinear force equation, F = −α x + β x^3. The user initially integrates the force to find the work done when the spring is stretched from its equilibrium position to 0.15 m, resulting in a negative value, which leads to confusion about the interpretation of work. It is clarified that the negative value indicates the work done by the spring as it resists being stretched, not the work done by the person pulling it. The final answer, converted to milli-joules, is -22.35 mJ, highlighting the importance of units in the calculation. The discussion emphasizes the distinction between work done by the spring and work done by an external force.
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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?
 
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Yeah, but the problem asks for the work done by the spring, not the work done by the person pulling it.
 
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
 
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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
 
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).
 
-22.35. Thanks a lot, I can't believe I didn't see that.
 
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