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Work in a Spring

  • #1
I have a quick question:

10. When a 500-g mass is suspended from a spring, it stretches 2.5 cm.
What is the force constant of the spring? If you now pull the mass
down producing a further displacement of 5 cm, what is the work
done on the spring?


The answer for the second part the professor has:

W=.5kx^2 = .245J

How come we don't use the equation:
W=.5k(xf^2-xi^2) doesn't the spring already have a load on it that stretches it?
I got .49J
 

Answers and Replies

  • #2
275
2
x must always be measured from the equilibrium position - which means: whatever position you can leave the spring in, without it starting to oscillate.

If there was no mass on the spring, you would use the equation you gave.

Does that make sense?
 
  • #3
Doc Al
Mentor
44,882
1,129
don't neglect gravity

The answer for the second part the professor has:

W=.5kx^2 = .245J

How come we don't use the equation:
W=.5k(xf^2-xi^2) doesn't the spring already have a load on it that stretches it?
I got .49J
It's a bit tricky to understand.

When you use .5kx^2 with x measured from the equilibrium position, you are really finding the total potential energy change of the system including both spring PE plus gravitational PE. You are correct that the energy stored as spring PE is given by .5kx^2 where x is measured from the unstretched position of the spring. But what they are really asking for is the work you must do on the spring to stretch it an additional distance, not the total work done on the spring. (Gravity does part of the work.)

Let's work it out in detail and maybe it'll be clearer:

The amount of stretch in the spring when at equilibrium = mg/k
The spring PE at equilibrium = .5k(mg/k)^2
The amount of stretch in the spring when pulled an addition distance x = mg/k + x
The spring PE at that point = .5k(mg/k + x)^2

Measured from equilibrium, the change in spring PE = .5k(mg/k + x)^2 - .5k(mg/k)^2
Measured from equilibrium, the change in gravitational PE = -mgx

Measured from equilibrium, the total change in potential energy
= .5k(mg/k + x)^2 - .5k(mg/k)^2 -mgx = .5kx^2

Make sense?
 
  • #4
14
0
I understand it that way : (sorry if I repeat something you said Doc Al)

The string in the equilibrium position has a certain amount of elastic potential energy. We can take this equilibrium position's potential energy as a reference. You have to find the work done on the spring, so the difference of energy between its final and initial state. The elastic potential energy of a string is 1/2 * kx², mesured from the equilibrium position (of the system we study, here it's the loaded string equilibrium position). Here we have a new equilibrium position, and so the displacement is only 5cm. And so the initial energy (being our reference) is 0 and the final energy (relative to our reference) is 1/2 kx², where x=5cm.

I hope I helped you understand if it wasnt already the case with previous explanations.
 

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