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When a person walks (assuming the feet don't slip), the ground is pushed backward by the foot while the force of static friction opposes the force and the foot remains stationary while the upper body moves. - This is what I've understood. However, the upper body moves - I believe, since the foot exerts a force backward and the person moves forwards, the work done by the body must be negative, while, since friction acts along the direction of motion, the work done by it should be positive. These two negative and positive forces must cancel each other out, so the net work done while walking is zero.

Could someone please explain if this right or wrong?

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The person is pushing himself forwards, as it were (as it is). His feet push in a backward direction and he/she acquires a momentum forward (while the Earth receives a momentum backward).

Because of this pushing, the person delivers positive work.

The kinetic as well as potential energy stays constant during walking on a flat surface. To keep the kinetic energy constant though the walker will have to deliver work. The potential energy ($mgh$) stays, on average, constant. Energy is used by the walker though to cause or counteract the differences in potential energy. Every time the walker takes a step the height of his center of mass varies.

Every time he/she makes another step, he/she must push him/herself forward to maintain a constant momentum (kinetic energy). Because else, due to friction, her/his momentum would be getting smaller. For example, when walking against the wind, you have to keep using energy as well as working to keep your velocity (and thus momentum) constant.
Even in the ideal case of walking on a perfectly flat surface, and no sigh of the wind, you have to overcome the internal frictions inside your body.
You can even imagine that you are walking on a flat surface in outer space (you need some extra equipment though). In this case, too, you'll have to do work.

See also the answers to this question.

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  • $\begingroup$ This is only true while the walker is increasing speed or climbing a hill. $\endgroup$ – Peter Sep 28 at 12:33
  • $\begingroup$ @Peter Why? Because his momentum is constant while walking? $\endgroup$ – Deschele Schilder Sep 28 at 12:36
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    $\begingroup$ No, because his energy (kinetic plus potential) is constant. $\endgroup$ – Peter Sep 28 at 12:46
  • $\begingroup$ @Peter Every time he/she makes another step, he/she must push him/herself forward to maintain a constant momentum (kinetic energy). Because else, due to friction, her/his momentum would be getting smaller. The potential energy $mgh$ stays the same indeed (on the average). $\endgroup$ – Deschele Schilder Sep 28 at 12:47
  • $\begingroup$ friction does not always result in things becoming slower. $\endgroup$ – Peter Sep 28 at 13:30
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If a person is walking on a level path in a straight line at constant speed the total kinetic energy is constant, so no total work is done.

The external forces acting on the person are gravity downwards, the normal force of the ground pushing the person upwards, friction with the ground pushing the person forward, and air resistance pushing the person backwards.

In a simplified analysis we would assume the person does not move vertically at all. In this case friction and direction of travel are in the same direction so the force does positive work, while air resistance and direction of travel are in opposite directions so that force does negative work. Total is zero.

If we ignore air resistance so that friction is the only force, it does positive work when the person starts walking (increasing speed) and negative work when the person stops (decreasing speed). The total work done from start to end during the walk would again be zero.

In actual fact walking is a more complex activity than it appears. Start when the right leg is upright, the body is moving forwards and the left leg is moving past it. As the body moves forward it is also moving downward, and when the left (now the leading) heel hits the ground, it must first push backwards and upwards. At this time it is doing negative work. Next the right toes and instep push forward to lift the body over the left leg, doing (positive) work against gravity as the body rises. The cycle then repeats. The positive and negative work adds up to zero over several cycles.

We usually feel the force more as we drive forward with the back foot, though your legs will become tired as they absorb energy while you walk downhill. Even walking on level ground it doesn't feel as though no work is done because of the energy expended within the body as muscles contract and relax during the process.

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    $\begingroup$ That means the force of friction and air resistance must be the same, right? What if the air resistance is zero?, i.e a person is moving on earth in a vacuum chamber. $\endgroup$ – Yashas Sep 28 at 12:22
  • $\begingroup$ Then the friction force would also be zero. There is some variation as the walker tilts forwards or backwards, and as he rises or falls, but it would average to zero. $\endgroup$ – Peter Sep 28 at 12:30
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    $\begingroup$ If friction is zero , locomotion shouldn't be possible $\endgroup$ – Yashas Sep 28 at 13:43
  • $\begingroup$ If a person is walking on a level path in a straight line at constant speed the total kinetic energy is constant, so no total work is done. This is simply not true. As I already said in a comment to my answer. Well, it's true that the velocity is constant but this is caused by the work done. So the last part of what I cited here is clearly not true. $\endgroup$ – Deschele Schilder Sep 28 at 13:45
  • $\begingroup$ @DescheleSchilder, if velocity is constant, kinetic energy is constant. If path is level potential energy is constant (more or less). Therefore total work done is zero. Any positive work on the walker must be balanced by negative work on the walker. $\endgroup$ – Peter Sep 28 at 13:50

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