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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?

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?