# Potential Energy on an ideal spring

• clope023
In summary, when a 3.15kg weight is hung from an ideal spring of negligible mass, its length changes from 12cm to 13.3cm. The potential energy of the spring is 28.22J.
clope023
[SOLVED] Potential Energy on an ideal spring

## Homework Statement

An ideal spring of negligible mass is 12.00cm long when nothing is attached to it. When you hang a 3.15kg weight from it, you measure its length to be 13.30cm.

If you wanted to store 10.0J of potential energy in this spring, what would be its total length? Assume that it continues to obey Hooke's law. (also says there might be 2 answers)

## Homework Equations

K1 + U1 + Wother
= K2 + U2

or

Wel = 1/2ky1^2 - 1/2ky2^2
= Uel1 - Uel2

U = mgy

## The Attempt at a Solution

12cm = .12m = y1
13.30cm = .133m = y2
10J = Wother

1/2ky1^2 = 10J + mgy1 + 1/2ky2^2

1/2k(.0144m^2) = 10J + (3.15kg)(9.8m/s^2)(.133m) + 1/2k(.0177m^2)

-14.11J = 1/2k(.0177m^2-.0144m^2)

-14.11J = 1/2k(.0033m^2)

-28.22J = k(.0033m^2)

-28.22m = (.003m^2)

is this right? would I just divide -28.22m by .003m^2 and get a #? I'm pretty sure it's not right but I'm not sure of other equations to use, should I use kinematics as well?

edit: or perhaps I should have another potential energy in there for y1? or use the potential energy for mass in y1 and solve for y2?

or perhaps use this equation?

K1 + Ugrav1 + Uel1 + Wother = K2 + Ugrav2 + Uel2

seems like this would work better than what I was using?

Last edited:
The potential energy of a spring is given by:

$$U=\frac{1}{2}kx^2$$

You can work out the spring constant from the initial info they give you.

Kurdt said:
The potential energy of a spring is given by:

$$U=\frac{1}{2}kx^2$$

You can work out the spring constant from the initial info they give you.

you saying I should do:

10J = 1/2k((.133m)^2-(.12m)^2))

20J = k(.0033m^2)

20J/.003m^2 = k

k = 6060?

or should I use the gravitational potentiona energy from the 3.15kg mass?

(3.15)gy = 1/2k((.133m)^2-(.12m)^2))?

You need to work out the spring constant by how far the mass stretches the spring, then use the potential energy formula I gave to solve for x. Since the answer will involve a square root there will be two answers.

Kurdt said:
You need to work out the spring constant by how far the mass stretches the spring, then use the potential energy formula I gave to solve for x. Since the answer will involve a square root there will be two answers.

so:

10J = 1/2k(.133m)^2

20J = k(.0177m^2)

20J/.0177m^2 = k

1129.9 = k

and mgy = 1/2kx^2

(3.15(9.8)(.133m)2/k)^1/2 = x?

x = -8.5cm, 8.5cm ?

The spring constant is a force per unit length. You can work it out by dividing the weight of the mass by how much the spring moves from equilibrium. Then solve for x in the potential energy equation.

Kurdt said:
The spring constant is a force per unit length. You can work it out by dividing the weight of the mass by how much the spring moves from equilibrium. Then solve for x in the potential energy equation.

so k = (mg)/y2 = (13.5*9.8)/.133m = 232J?

and is the potential energy in the equation you gave gravitational or elastic?

cause I could solve it like this

mgy = 1/2kx^2

4.11J = 1/2kx^2

8.22J = kx^2

8.22J/232J = x^2

.035 = x^2

x = -.188m, .188m = -19cm, 19cm?\

or is it:

10J = 1/2kx^2

20J = kx^2

20J/232J = x^2

.086 = x^2

x = -.29m, .29m = -29cm, 29cm?

Last edited:
clope023 said:
so k = (mg)/y2 = (13.5*9.8)/.133m = 232J?

No, the spring stretches 1.3cm from equilibrium. The units of the spring constant are not joules. The only purpose of the information about the mass is so you can determine the spring constant.

clope023 said:
and is the potential energy in the equation you gave gravitational or elastic?

Its the elastic potential energy. That is the energy stored in the spring.

Now you know how to obtain the spring constant and you know how much energy you want to store you can solve for x in the potential energy equation.

Kurdt said:
No, the spring stretches 1.3cm from equilibrium. The units of the spring constant are not joules. The only purpose of the information about the mass is so you can determine the spring constant.

Its the elastic potential energy. That is the energy stored in the spring.

Now you know how to obtain the spring constant and you know how much energy you want to store you can solve for x in the potential energy equation.

ooooooooh, it goes from 12cm to 13.3cm which is 1.3cm which .013m, so

Fx = kx, k = Fx/x = = (3.15kg*9.8m/s^2)/.013m = 30.9N/.013m = 2377N/m

and 10J = 1/2kx^2

20J = kx^2

20J/(2377N/m) = x^2

x = .092m = -9.2cm, 9.2cm?

Thats good but they want the total length. So that will be 12cm +/- 9.2.

Kurdt said:
Thats good but they want the total length. So that will be 12cm +/- 9.2.

right the initial length when nothing was attached to it, so it would be:

2.8, 21.2? thanks alot!

## What is potential energy?

Potential energy is the energy that an object has due to its position or state. It is stored energy that can be converted into other forms of energy.

## What is an ideal spring?

An ideal spring is a theoretical concept in physics where a spring is assumed to have no mass, no friction, and obeys Hooke's Law perfectly. This means that the force exerted by the spring is directly proportional to the displacement from its equilibrium position.

## How does potential energy relate to an ideal spring?

Potential energy on an ideal spring is directly related to the displacement of the spring from its equilibrium position. The more the spring is stretched or compressed, the greater the potential energy it has.

## What is the formula for calculating potential energy on an ideal spring?

The formula for calculating potential energy on an ideal spring is: PE = 1/2 * k * x^2, where PE is the potential energy, k is the spring constant, and x is the displacement from equilibrium.

## What factors affect the potential energy of an ideal spring?

The potential energy of an ideal spring is affected by the spring constant, which is determined by the material and shape of the spring, and the displacement from equilibrium. The greater the spring constant or displacement, the greater the potential energy.

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