tun1ju2k1ki

2022-10-08

Suppose a mass of $M$ kg is hanging from a spring in earth. The mass will stretch the spring about x m. So the change in the gravitational potential energy is $mgx$ J (supposing x to be very small compared to the radius of earth).

And this amount of energy will be stored in the spring as potential energy. So,

Change of gavitational energy = $mgx$ = potential energy stored in the spring

And it seems that the potential energy stored in a spring is proportional to displacement x. But the potential energy in a spring is $U=\frac{1}{2}k{x}^{2}$ and so it's proportional to ${x}^{2}$, the square of displacement. So surely I am wrong somewhere. But where am I wrong?

And this amount of energy will be stored in the spring as potential energy. So,

Change of gavitational energy = $mgx$ = potential energy stored in the spring

And it seems that the potential energy stored in a spring is proportional to displacement x. But the potential energy in a spring is $U=\frac{1}{2}k{x}^{2}$ and so it's proportional to ${x}^{2}$, the square of displacement. So surely I am wrong somewhere. But where am I wrong?

procjenomuj

Beginner2022-10-09Added 8 answers

Imagine that when you attach the mass to the unstretched spring you are holding the mass in you hand. You then gently lower the mass until the uplift $k{x}_{\mathrm{m}\mathrm{a}\mathrm{x}}$ of stretched spring just balances the dowward weight mg of the mass. Gravity has done work to stretch the spring, but the net downward force at stretch x

$F(x)=mg-kx$

has also done work

${\int}_{0}^{{x}_{\mathrm{m}\mathrm{a}\mathrm{x}}}(mg-kx)dx=mg{x}_{\mathrm{m}\mathrm{a}\mathrm{x}}-\frac{1}{2}k{x}_{max}^{2}$

on your hand. Thus the difference in your two formulae

$mg{x}_{\mathrm{m}\mathrm{a}\mathrm{x}}-\frac{1}{2}k{x}_{\mathrm{m}\mathrm{a}\mathrm{x}}^{2}$

is accounted for by the work done on you.

If you just attatch and let go, the mass would bounce up and down and ther would also be kinetic energy to keep track of.

$F(x)=mg-kx$

has also done work

${\int}_{0}^{{x}_{\mathrm{m}\mathrm{a}\mathrm{x}}}(mg-kx)dx=mg{x}_{\mathrm{m}\mathrm{a}\mathrm{x}}-\frac{1}{2}k{x}_{max}^{2}$

on your hand. Thus the difference in your two formulae

$mg{x}_{\mathrm{m}\mathrm{a}\mathrm{x}}-\frac{1}{2}k{x}_{\mathrm{m}\mathrm{a}\mathrm{x}}^{2}$

is accounted for by the work done on you.

If you just attatch and let go, the mass would bounce up and down and ther would also be kinetic energy to keep track of.

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?

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?

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