# Work, spring, mass, gravity -part 2

• david316
In summary, the conversation revolved around a thought experiment involving a mass hanging from a spring and the concept of work and energy conservation. The first experiment involved lifting the mass 1m and the second experiment involved pulling the mass down 1m. It was discussed that the amount of work done depends on the difference between the help received (from the spring or gravity) and the force working against it. Ultimately, it was concluded that the amount of work done is the same in both experiments, regardless of whether it is lifting or pulling the mass. The conversation also touched on the computation of work and the relevance of gravitational potential energy and strain in the spring.

#### 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 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?

david316 said:
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.

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.

Not sure if my last post is correct. Any help would be much appreaciated.

david316 said:
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.

Last edited:
Great. Thanks

## 1. What is the relationship between work and mass in a spring system?

The amount of work done on a spring is directly proportional to the mass attached to it. This means that as the mass increases, the amount of work required to stretch or compress the spring also increases.

## 2. How does gravity affect the motion of a spring system?

Gravity plays a significant role in the motion of a spring system. It acts as a restoring force, pulling the mass towards the equilibrium point and causing the spring to oscillate. The strength of gravity also influences the frequency and amplitude of the oscillations.

## 3. What factors can affect the spring constant in a mass-spring system?

The spring constant, also known as the stiffness of the spring, can be affected by various factors such as the material and shape of the spring, the number of coils, and the amount of tension or compression applied to the spring.

## 4. How does the work done on a spring change when the system is in a different gravitational field?

The work done on a spring will vary depending on the strength of the gravitational field. In a higher gravitational field, more work is required to stretch or compress the spring compared to a lower gravitational field.

## 5. Can a spring system experience resonance?

Yes, a spring-mass system can experience resonance when the frequency of the external force applied to the system matches the natural frequency of the spring. This results in a significant increase in amplitude and can potentially damage the system if the force is too strong.