# Potential energy of a displaced mass on a spring

In summary, to calculate the potential energy of a spring with a restoring force of (-k1x)+(k2x^2), one can use the equation U=½kx^2. However, this equation is only valid if the force is linear, so for this particular case, the correct way to calculate potential energy is ΔU=∫(-k1x)+(k2x^2)dx. This will result in the equation ΔU=-k1½x0^2+k2⅓x0^3, where ΔU represents the difference in potential energy between two states, and x0 represents the displacement from the initial state of x=0.

## 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.

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?

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

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

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

 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

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.

## 1. What is potential energy?

Potential energy is the stored energy an object possesses due to its position, shape, or condition. It is the energy that an object has the potential to convert into other forms of energy.

## 2. What is the potential energy of a displaced mass on a spring?

The potential energy of a displaced mass on a spring is the energy stored in the spring as a result of the displacement of the mass from its equilibrium position. It is also known as elastic potential energy.

## 3. How is the potential energy of a displaced mass on a spring calculated?

The potential energy of a displaced mass on a spring can be calculated using the formula PE = 0.5 * k * x^2, where PE is the potential energy, k is the spring constant, and x is the displacement of the mass from its equilibrium position.

## 4. What factors affect the potential energy of a displaced mass on a spring?

The potential energy of a displaced mass on a spring is affected by the mass of the object, the spring constant, and the displacement from the equilibrium position. The potential energy increases with an increase in mass and displacement, and with a decrease in spring constant.

## 5. How does the potential energy of a displaced mass on a spring relate to the Law of Conservation of Energy?

The potential energy of a displaced mass on a spring is an example of potential energy, which is one of the two main forms of energy described by the Law of Conservation of Energy. This law states that energy cannot be created or destroyed, only transferred or converted from one form to another. Therefore, the potential energy of a displaced mass on a spring can be converted into kinetic energy or other forms of energy, but the total energy remains constant.

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