# Spring constant based on change in potential energy

• sanctifyd83
In summary, the problem involves a compressed spring with 138 J of energy stored in it and when it is decompressed by 125 cm, it now stores 82.6 J of energy. The goal is to find the spring constant. The equations used are ΔU = 1/2kxf2 - 1/2kxi2, U(x) = 1/2kx2, U1 = 1/2kx12, and U2 = 1/2k(x1 + 1.25 m)2. After solving for x1, it is found to be -5.5 m, which leads to a value of 92000 N/m for the spring constant.
sanctifyd83

## Homework Statement

A compressed spring has 138 J of energy stored in it. When it is decompressed by 125 cm, it now stores 82.6 J of energy. What is the spring constant

## Homework Equations

General:
ΔU = 1/2kxf2 - 1/2kxi2
U(x) = 1/2kx2
U1 = 1/2kx12
U2 = 1/2k(x1 + 1.25 m)2

## The Attempt at a Solution

U1 = 1/2kx12

k = 2U1/x12

U2 = 1/2 (2U1/x12)(x1 + 1.25 m)2

U2 = U1/x12 (x12 + 2.5m x1 + 1.5625 m2

U2 = U1 + (2.5 m x1/x1) + (1.5625 m2 U1/x12)

x12 ΔU = (2.5m U1/x1 + 1.5625m2 U1/x12) x12

ΔU x12 = 2.5m U1 x1 + 1.5625m2 U1

Substitute values and get them all to one side:
55.4J x12 - 345Jm x1 - 215.625Jm2 = 0

x1 = 6.8 cm and -.57 cm

138J = 1/2k (.068 m )2

276J/.00462 m2 = 59740 N/m = k? Obviously that can't be right.
Any help on this would be great!

sanctifyd83 said:

U2 = U1 + (2.5 m x1/x1) + (1.5625 m2 U1/x12)

This equationis wrong; the middle term does not have units of energy.

I continue to recommend that people leave numbers out until the end, and that they check dimensions term-by-term everywhere.

Last edited:
Hello sanctifyd83. Welcome to PF!

sanctifyd83 said:
ΔU x12 = 2.5m U1 x1 + 1.5625m2 U1

Substitute values and get them all to one side:
55.4J x12 - 345Jm x1 - 215.625Jm2 = 0

Is ΔU positive or negative?

rude man said:
This equationis wrong; the middle term does not have units of energy.

I continue to recommend that people leave numbers out until the end, and that they check dimensions term-by-term everywhere.

Thank you, rude man. Yes I input the middle term incorrectly. It should be 2.5 m/x1. I did it correctly when I worked it out though, I don't believe it affects the rest of my "solution."

TSny said:
Hello sanctifyd83. Welcome to PF!

Is ΔU positive or negative?

Thank you for that. Yes, the value ought to be negative. This gives the value for x1 slightly different negative values which, when squared, become positive. The result is still an extremely large value for k.

x1 would equal -5.5 or -.7
Using -5.5 k would be around 90000N/m and -.7 would give an even larger number.

I agree with the -5.5 m. (You should see a reason why you have to throw out -0.7 m)

But I don't see how you're getting such a large value for k. Can you show how the x1 = -5.5 m leads to k = 90000 N/m?

TSny said:
I agree with the -5.5 m. (You should see a reason why you have to throw out -0.7 m)

But I don't see how you're getting such a large value for k. Can you show how the x1 = -5.5 m leads to k = 90000 N/m?

Forgive me as I'm trying to understand the material but I think that x1[/SUB's units would be in cm? If this is the case -5.5 cm turns into -.055 m which when squared becomes .003 m2. 276J/.003 m2 = 92000N/m.
How do you know what the variable's initial units are? I assumed we were working with it in cm since the problem initially had the length units in cm. I know that I converted it to meters but it seems that the quadratic function canceled all units except the ones for x1.

The energy is given in Joules, an SI unit. Thus, distances should be in meters. So, it's good that you converted the 125 cm to meters. When you solve the equations, the distances will still be in meters.

EDIT: Take your equation 55.4J x12 + 345Jm x1 + 215.625Jm2 = 0

The Joules cancel out. Can you then see by inspection of the remaining units that x1 must have units of meters?

Last edited:
That absolutely makes sense, I couldn't quite grasp that. It seemed like I was just arbitrarily inputting units. Thanks so much.

Maybe I'm missing something:
$$138 = \frac{1}{2}kx_0^2$$
$$82.6 = \frac{1}{2}k(x_0-1.25)^2$$
Divide one equation by the other to get x0 (the k's cencel). Then find k from the first equation.

Chet

2 people
I'm with you Chestermiller

Chestermiller said:
Maybe I'm missing something:
$$138 = \frac{1}{2}kx_0^2$$
$$82.6 = \frac{1}{2}k(x_0-1.25)^2$$
Divide one equation by the other to get x0 (the k's cencel). Then find k from the first equation.

Chet

Yes, this is much nicer. The OP had already set up and worked the problem correctly except for a sign error, so I didn't want to start over.

But it's certainly worth while to point out a better method of solution, and I should have done that.

Thanks.

Thanks so much everyone. It helped out a lot.

## 1. What is the definition of a spring constant?

A spring constant, also known as a force constant, is a physical property that describes the stiffness of a spring. It is a measure of the amount of force required to stretch or compress a spring by a certain distance.

## 2. How is the spring constant related to the change in potential energy?

The spring constant is directly proportional to the change in potential energy. This means that as the spring constant increases, so does the amount of potential energy stored in the spring when it is stretched or compressed.

## 3. What factors affect the spring constant?

The spring constant is affected by three main factors: the material of the spring, the cross-sectional area of the spring, and the length of the spring. Generally, stiffer materials, larger cross-sectional areas, and shorter lengths result in higher spring constants.

## 4. How is the spring constant measured?

The spring constant can be measured by applying a known force to a spring and measuring the resulting displacement. By using Hooke's Law, which states that the force applied to a spring is directly proportional to the amount of displacement, the spring constant can be calculated.

## 5. Why is the spring constant important?

The spring constant is an important concept in physics and engineering because it helps us understand the relationship between force and displacement in springs. It is also used in various applications, such as designing suspension systems and creating accurate force measuring devices.

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