# Differculty working out what is doing work in a mass attached to a spring system

1. Sep 4, 2012

### david316

Hello,

I am trying to get an understanding of work and energy conservation and am getting confused.

My thought experiment revolves around a mass hanging from a spring and appling a force to the resting mass.

Assume a 10kg mass hanging from a spring with k = 30 N/m. The mass will stretch the spring to x1 ~= -100/30 = -3.3m (-ve x is downwards). Given an equilbrium position of x1 = -3.3m I then "lift" the mass by applying an upwards force of 50N.

The total force acting on the mass will be 100N (due gravity) minus 50 N due to me lifting the spring. Therefore the mass will come to rest at x2 = -50/30 = 1.67m.

By lifting the mass I have done work on the spring by adding PE to the mass.
ΔPE to mass = mgΔh = 100*1.67 = 167J

I have also removed PE from the spring as I have moved it back towards its unstrained position.
Δ PE to spring = 0.5×(x2^2 - x1^2) = -125J

Does this mean the total work done by me lifting the mass is 167-125 = 42J
Or do I just use W = f×d = 50 x 1.67 = 83.5J
Or have I made a mistake somewhere? Do I need to account for the potential for the mass to oscillate?

Any help would be great!!

2. Sep 4, 2012

### Staff: Mentor

If you apply a force of 50N to the mass over the distance 1.67m, the mass won't be at rest after it's been lifted 1.67 meters. It will be moving rapidly upwards, with a bunch of extra kinetic energy and will overshoot the new equilibrium point by a fair amount.

Consider the situation before you apply the upwards force. If you apply even the most minuscule amount of force, just a tiny fraction of 50N, the mass will move up just a bit. The spring tension balanced the force of gravity; you added an additional upwards force so now the net force on the mass is upwards and it starts to accelerate upwards. The only reason it doesn't keep on moving upwards (as it would if you were using a counterweight instead of a spring) is that as it moves up the spring shortens so is exerting less upwards force and it reaches a new equilibrium. Increase the force a bit more, and you'll get it a bit further up... and so one until you're applying a force of 50N after lifting it 1.67 meters.

So it doesn't take 50N over the entire 1.67 distance to get the object to the new equilibrium point, and if you do apply 50N over that entire distance... You're throwing it upwards, past the equilibrium point.

Last edited: Sep 4, 2012
3. Sep 4, 2012

### henrique_p

Yeah, I think that's the point.

4. Sep 4, 2012

### david316

Ok. So if I very slowly "lifted" the mass to the position x2 = -1.67m, such that the forces balanced and it was at rest at its new equilibrium position, I could say I would have done 42J work and would be appling 50N upwards at that point?

5. Sep 4, 2012

### henrique_p

I just don't understand something in your problem: when you said you put up a Force of 50N , this is the Resultant Force (150N [yours] - 100N[m.g] = 50N [Resultant Force]), or it is the Force you are applying? Because if you apply a Force of 50N to an object under 100N vertically, it won't move.

Also, I calculated the Elastic Potential Energy before lifting the mass (with the -3.3m stretch), and I found 163.35J stored in the spring.

6. Sep 4, 2012

7. Sep 4, 2012

### Staff: Mentor

yes. The force you'd be be applying at the start of the move would be 0N and and at the end it would be 50N.

To really do it right, you'd have to integrate $$W=\int{F(x)dx}$$ from 0 to 1.67, where F(x) is the force required to support the weight with the help of the spring after it's been lifted by an amount x.

But if you have a perfect spring, as this problem assumes, it turns out that the value of this integral is just $\frac{XF(X)}{2}=\frac{Fd}{2}$ where X is the total amount you've lifted the thing. And that's why you were off by a factor of two when you calculated W=Fd.

8. Sep 4, 2012

### david316

Thanks Nugatory. That makes sense.