Work, spring, mass, gravity -part 2

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Main Question or Discussion Point

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 moving 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 = -3.33m (-ve x is downwards). Given an equilbrium position of x1 = -3.33m I then "lift" the mass 1m.

By lifting the mass I have done work on the spring by adding PE to the mass.
ΔPE to mass = mgΔh = 100*1 = 100J

I have also removed PE from the spring as I have moved it back towards its unstrained position.
ΔPE to spring = 0.5×30x((-2.33)^2 - (-3.33)^2) = -85J

I assume this mean the work I have done by lifting the mass is 100 - 85 = 15J ?

I then allow the mass to return to the -3.33m position for experiment 2. In this experiment I pull down on the mass until it is at the -4.33m position.

By pulling the mass downwards I have removed PE from the mass.
ΔPE removed from mass = mgΔh = 100*1 = -100J

I have also added PE to the spring as I have stretched it.
ΔPE added to spring = 0.5×30x((-4.33)^2 - (-3.33)^2) = 115J

I assume this mean the work I have done by pulling down the mass is 100 - 85 = 15J which is the same a lifting the mass?

So working with or against gravity doesn't change the amount of work I do? This doesn't seem intuitive to me?
 

Answers and Replies

  • #2
Nugatory
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Hello,
So working with or against gravity doesn't change the amount of work I do? This doesn't seem intuitive to me?
When the mass is in its initial equilibrium position, anything you to do move the mass up or down will change the gravitational potential energy up or down and change the spring potential energy the other way. So one way gravity is helping you fight the spring, and the other way the spring is helping you fight gravity.

The amount of work that you do depends on the difference between the help that you get (from the spring when lifting against gravity, from gravity when pulling down against the spring) and the force that you're working against, and it's just coincidence that you chose a second thought experiment that leads to the exact same amount of work as the first. Consider how much work you'd have to do to pull the mass down by 500 meters; now the spring will be pulling up much more than gravity is pulling down, and you'll be doing a huge amount of work even though gravity is on your side.
 
  • #3
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Working through the 500m doesn't seem to make any difference....

Assume a 10kg mass hanging from a spring with k = 30 N/m. The mass will stretch the spring to x1 ~= -100/30 = -3.33m (-ve x is downwards). Given an equilbrium position of x1 = -3.33m I then "lift" the mass 500m.

By lifting the mass I have done work on the mass by adding PE to the mass.
ΔPE to mass = mgΔh = 100*500 = 50000J

I have also added PE to the spring as I have moved it back and past its unstrained position.
ΔPE to spring = 0.5×30x((496.67)^2 - (-3.33)^2) = 3700000J

I have done 3750000J work by lifting the mass?

I then allow the mass to return to the -3.33m position for experiment 2. In this experiment I pull down 500m.

By pulling the mass downwards I have removed PE from the mass.
ΔPE removed from mass = mgΔh = -50000J

I have also added PE to the spring as I have stretched it.
ΔPE added to spring = 0.5×30x((-503.33)^2 - (-3.33)^2) = 3800000J

I assume this mean the work I have done by pulling down the mass is 37500000 which is the same as lifting the mass.
 
  • #4
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Not sure if my last post is correct. Any help would be much appreaciated.
 
  • #5
jbriggs444
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Working through the 500m doesn't seem to make any difference....

Assume a 10kg mass hanging from a spring with k = 30 N/m. The mass will stretch the spring to x1 ~= -100/30 = -3.33m (-ve x is downwards). Given an equilbrium position of x1 = -3.33m I then "lift" the mass 500m.

By lifting the mass I have done work on the mass by adding PE to the mass.
ΔPE to mass = mgΔh = 100*500 = 50000J
So rather than compute the work you have done as the path integral of force dot incremental displacement, you are computing it by looking at the resulting change in PE. That's fine.

And I agree that the change in gravitational PE is +50,000 J [assuming G = 10 m/sec^2]

I have also added PE to the spring as I have moved it back and past its unstrained position.
ΔPE to spring = 0.5×30x((496.67)^2 - (-3.33)^2) = 3700000J
This looks correct. You've relieved 3.33 meters worth of strain and added 496.67 meters worth of strain and 1/2 k d^2 would be the relevant formula.

I have done 3750000J work by lifting the mass?
Yep.

I then allow the mass to return to the -3.33m position for experiment 2. In this experiment I pull down 500m.

By pulling the mass downwards I have removed PE from the mass.
ΔPE removed from mass = mgΔh = -50000J

I have also added PE to the spring as I have stretched it.
ΔPE added to spring = 0.5×30x((-503.33)^2 - (-3.33)^2) = 3800000J

I assume this mean the work I have done by pulling down the mass is 37500000 which is the same as lifting the mass.
Yes. And this makes perfect sense.

Close your eyes for a moment and ignore the fact that there are springs and gravity and what-not involved. All you have is a black box with a doo-hickey sticking out.

This doo-hickey has an equilibrium position. If you push it one meter North of the equilibrium position it resists with 30 Newtons of force. If you push it two meters, it resists with 60 Newtons. Similarly if you push it one or two meters south.

Pushing this doo-hickey 500 meters North clearly imparts the same amount of work as pushing it 500 meters South.
 
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  • #6
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Great. Thanks
 

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