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Suppose a man of mass 50 kg climbs a stair of height 10 m.Clearly it gains 5000 J of potential energy.But since work done by normal reaction on the man is zero does'nt it violates the principle of conservation of mechanical energy because no non-conservative force is acting on the man and thus mechanical energy should be conserved.

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  • $\begingroup$ But since work done by normal on the man is zero - that seems wrong. Assuming by normal you mean the vertical gravitational force the work the man does against it is equal to $mgh$. $\endgroup$ Oct 17, 2017 at 16:20
  • $\begingroup$ By normal I mean normal reaction exerted by stairs on the man $\endgroup$ Oct 17, 2017 at 16:23
  • $\begingroup$ Hello Gagandeep I need your help with this question. physics.stackexchange.com/q/364216/171832 $\endgroup$
    – Divx
    Oct 21, 2017 at 18:15

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By "work done by normal", you probably mean that since the foot does not move relative to the stairs, whatever the normal force on man exerted by the stairs, there cannot be work done by this force as displacement is zero. In the case of discrete stairs with horizontal top surface, that is correct.

However, when your muscles contract to climb each stair, energy is spent. There is no such thing as "conservation of mechanical energy". All that is required is that total energy is conserved.

Even with mechanical energy, suppose you have a compressed spring with two masses, one at each end: O///\O and release the spring. The balls will gain speed and have kinetic energy, even if there was no outside work performed on the total "spring plus balls" system. Energy stored in the spring was used to push the balls.

In your example, chemical energy is used to move your muscles in a very complex manner that results in a gain of gravitational potential energy. You could get the same result by having a complex mechanism that uses energy stored in a compressed spring to perform each step.

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  • $\begingroup$ Ok, That makes a lot of sense.But my textbook states that total mechanical energy of a body is conserved when no non-conservative force acts on the body.Is that statement flawed or I am missing something? $\endgroup$ Oct 17, 2017 at 16:44
  • $\begingroup$ That is correct in the context of mechanics. However, there are many situations in which mechanical energy is not conserved. You can read more on energy in the Feynman lectures on physics: feynmanlectures.caltech.edu/I_04.html. $\endgroup$ Oct 17, 2017 at 16:59
  • $\begingroup$ And I forgot, some more reading on mechanical energy: feynmanlectures.caltech.edu/I_13.html and feynmanlectures.caltech.edu/I_14.html $\endgroup$ Oct 17, 2017 at 17:10
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When the man applies force to the stairs, the stairs apply an equal force back.

The mans force down is acting on the stairs. If the stairs started moving downwards, you could say the man is doing work on the stairs. Instead, the stairs do not move; the force is static, and there is no work done on the stairs.

The stairs provide an equal and opposite reaction upwards; but the man is moving, so there is work done on the man by the stairs.

By pushing down with a force greater than his own weight, the man is able to use the stairs to generate an upwards push. The energy for the push doesn't come from the stairs though. It comes from the muscles, which get their energy through food (chemical energy).

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  • $\begingroup$ "...the stairs do not move" And, neither does the foot that is standing on the stair, so the stair does no work on the foot. If the stair does no work on the only part of the man that touches it, then the stair does no work on the man. $\endgroup$ Oct 17, 2017 at 21:15
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    $\begingroup$ @jameslarge Except the man is able to use his legs to raise the centre of mass of his entire body, which is movement. If you isolate the stairs on the other hand; the forces result in no net movement (assuming properly installed stairs). I was going to get into the spiral of the temporary work done in the bending of the stairs as well; but I thought that might be a bit much. You can argue that the work done on the man is done by the man though, that is true. The man has to do work on the stairs before they do work onto him, so it's not like the stairs push him unprompted. $\endgroup$
    – JMac
    Oct 17, 2017 at 21:29
  • $\begingroup$ Work is a transfer of energy. If I ride up an escalator, its motor transforms electric energy into mechanical energy (i.e., work), and it does work on me, lifting me to the upper level of the shopping mall. But if I take the stairs instead, where does the energy come from? It doesn't come from the stair. It comes from my body. If the stair is not transferring energy to me, then it is not doing work on me. "Energy" and "work" are two different names for the same thing. $\endgroup$ Oct 18, 2017 at 13:31
  • $\begingroup$ @jameslarge The work ultimately comes from yourself; but you cannot directly do the work on yourself. First you need to do work on the stairs, giving them an elastic potential energy when they react to your step. Then, when you finish applying the force, the stairs are able to provide the energy back as they go back to their original shape. $\endgroup$
    – JMac
    Oct 18, 2017 at 14:27
  • $\begingroup$ Elasticity is a real thing, but it is not needed in an explanation of how a person climbs stairs. $\endgroup$ Oct 18, 2017 at 15:33
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the work done by normal force is not zero. since the displacement of man and normal force is in same direction and work done by a force is integral f.dx the work done by normal would be normal force*displacement in height. since man is applying external force to increase his mechanical energy so mechanical energy won't remain constant. mechanical energy of only those object which are isolated from other mass system and no external mechanical energy is provided remain constant.

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    $\begingroup$ The normal force acts through a zero displacement. It therefore performs zero work. $\endgroup$ Dec 3, 2018 at 9:54
  • $\begingroup$ @Emilio Pisanty when you place an object on the elevator and lift it through the height then it is the normal force only on the object that perform work of normal force * height raised. so work done by normal is not always zero it depends on the direction of displacement. if displacement is at right angle to the normal force then only work done by normal is zero since work is a dot or scalar product of force and displacement. $\endgroup$ Dec 3, 2018 at 10:17
  • $\begingroup$ This is incorrect. You're confusing the notion of work with that of center of mass work, e.g. see here. $\endgroup$
    – knzhou
    Dec 3, 2018 at 14:03

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