Potential energy of a displaced mass on a spring

[AsadaShino92]

Homework Statement

A spring of negligible mass exerts a restoring force on a point mass M given by F(x)= (-k1x)+(k2x^2) where k1 and k2 >0. Calculate the potential energy U(x) stored in the spring for a displacement x. Take U=0 at x=0.

Homework Equations

ΔU=∫F(x)dx
U=½kx^2

The Attempt at a Solution

Using the equation above I tried to find the potential energy of the spring after it has been displaced by some distance x. I integrated F(x) from x=0 to some displacement x=x0.

ΔU=∫(-k1x)+(k2x^2)dx

ΔU=-k1½x0^2+k2⅓x0^3

Is this the right way in finding the potential energy?

 

[BvU]

Almost. Basically you are going to refine your second relevant equation. So if $k_2 = 0$ you should find it back. Do you ?

[edit] hint: what about the discrepancy between 1st and 2nd relevant equation ?

 

[AsadaShino92]

Almost. Basically you are going to refine your second relevant equation. So if $k_2 = 0$ you should find it back. Do you ?

[edit] hint: what about the discrepancy between 1st and 2nd relevant equation ?

So then I can use the fact that ΔU= Uf-Ui= ½kxf^2-½kxi^2. Where f is final and i is initial?
If this is correct, then xi=0 and that term drops out. Then I would be left with ½kxf^2 = -k1½x0^2+k2⅓x0^3

 

[haruspex]

U=½kx^2

No. As BvU was hinting, that equation is only valid if F(x)=-kx.

So then I can use the fact that ΔU= Uf-Ui= ½kxf^2-½kxi^2.

No, for the reason given above.

ΔU=-k1½x0^2+k2⅓x0^3

Is this the right way in finding the potential energy?

Yes, except that you have a Δ on the left, implying a difference between two different states. On the right, you have assumed that one of those states is x=0. Try to write it consistently.