Spring constant based on change in potential energy

[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 = 1/2kx_f² - 1/2kx_i²
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

Quadratic formula gives:
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!

 

[rude man]

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.

 

[TSny]

Hello sanctifyd83. Welcome to PF!

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?

 

[sanctifyd83]

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."

 

[sanctifyd83]

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.

 

[TSny]

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?

 

[sanctifyd83]

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 x_{1[/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.}

 

[TSny]

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?

 

[sanctifyd83]

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

 

[Chestermiller]

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

Chet

 

[barryj]

I'm with you Chestermiller

 

[TSny]

Maybe I'm missing something:
[tex]138 = \frac{1}{2}kx_0^2[/tex]
[tex]82.6 = \frac{1}{2}k(x_0-1.25)^2[/tex]
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.

 

[sanctifyd83]

Thanks so much everyone. It helped out a lot.