If a Woman weighing $600 N$ steps on a bathroom scale containing a stiff spring, the spring is compressed $1.0 cm$ under her weight, the spring constant is $60000 N/m$.It turns out that the work done on the spring is $\frac{1}{2}kx^2=\frac{1}{2}(60000N/m)((0.01)(0.01))=3$ Joules. BUT isn't the work done on the spring $600N$ times $0.01m=6$ joules? Since the woman exerts a constant $600N$ force on the spring throughout the $0.01m$ displacement? Thanks in advance.

  • $\begingroup$ Can I know the reason for downvote? $\endgroup$ – Shannu Shannu Jan 10 '19 at 6:00
  • $\begingroup$ Work being $F\times d$ only works for constant forces, a spring doesn't exert a constant force, it depends on the compression. $\endgroup$ – Triatticus Jan 10 '19 at 6:01
  • $\begingroup$ @Triatticus The OP is asking about the work done by gravity that is assumed to be constant close to the Earth's surface, and why this does not match the work done by the spring, which they do not use $F\times d$ for $\endgroup$ – BioPhysicist Jan 10 '19 at 6:03
  • $\begingroup$ I see that, for some reason I thought the issue was the difference in work between a spring and gravity. $\endgroup$ – Triatticus Jan 11 '19 at 21:30

You are correct in saying that the work done by gravity is $6\ \rm J$. So where did the energy go?

Well consider the case where the woman is at rest above the uncompressed spring (magically hovering, I don't know) and then is released. Using energy conservation we find that $$\frac12kx^2=mgx$$ or $$x=\frac{2mg}{k}=2\ \rm{cm}$$

So we see that, if we had an ideal system of releasing the woman from rest above the spring she would actually sink $2\ \rm{cm}$ before coming to rest for the first time. But in your problem she only goes down by half this amount.

We must conclude that energy is lost due to dissipative forces. This makes sense. Without dissipative forces the woman would oscillate on the scale indefinitely. The dissipative forces are needed for the woman to come to a complete, lasting stop.

Therefore, gravity does $6\ \rm J$ of work. Half of it goes into the spring potential energy. The other half leaves the system through whatever mechanisms bring the woman to rest after stepping on the scale. (And of course this is all just focusing on if she starts right above the uncompressed spring at rest)

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  • $\begingroup$ But how is that integral magically encapsulating this fact? $\endgroup$ – Shannu Shannu Jan 10 '19 at 15:59
  • $\begingroup$ @ShannuShannu Which integral? I'm not sure I understand your question, can you explain more? $\endgroup$ – BioPhysicist Jan 10 '19 at 18:10
  • $\begingroup$ The work done on the spring by the constant 600N force is 6J,But if we calculate it using the formula 1/2x^2 we find it to be 3J,which is true.Since the work done on the spring is given by the integral of kxdx,which in turn gives us the formula 1/2kx^2,how is this working out? How is the derivation of this formula from that integral encapsulating your reasoning? I have actually asked a similar question here physics.stackexchange.com/q/453305/194502 $\endgroup$ – Shannu Shannu Jan 10 '19 at 18:45
  • $\begingroup$ @ShannuShannu It was a good idea to ask a new question, and the answer given there looks sufficient. I'm not sure there is much I can add there, unless the answer doesn't make sense to you. $\endgroup$ – BioPhysicist Jan 10 '19 at 20:49

The work provided is 6 J. But only part of this work is converted into elastic potential energy. The rest is converted into kinetic energy (the mass oscillates) and finally in heat. You would need a quasistatic transformation to convert all the work into elastic potential energy.

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  • $\begingroup$ Can you please elaborate on how the formula 1/2kx squared encapsulates these ideas.I mean the integral of force function with the distance dx spits us out 1/2kxsquared. $\endgroup$ – Shannu Shannu Jan 10 '19 at 5:43
  • $\begingroup$ To obtain the elastic potential energy, one writes $\delta {{W}_{ext}}={{F}_{op}}dx=kxdx$ and integrate. This supposes that the force exerted by the operator who compresses the spring is at every moment opposite to the action of the spring ${{F}_{op}}=-{{F}_{spring}}=+kx$. Quasistatic transformation: a continuous series of equilibrium states. $\endgroup$ – Vincent Fraticelli Jan 10 '19 at 6:04

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