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I'm going to ask a simple question, but it's bothering me. We all know that work done by a force which is perpendicular to its path of motion is zero.

Suppose I keep a chunk of mass in a perfectly frictionless sack, and I start moving from place A to place B. I am applying a force which is perfectly perpendicular to my way to place B, and I convinced gravity also to do so(assuming way from place A to B is perfectly straight trajectory, neglect curvature of earth).

Now as I think work done on the mass is zero, but who is carrying the mass from A to B? It can't move from A to B without any assistance, also at the same time my work done is zero!!!

My question is basically, who is doing the real work on the mass? If nobody is doing, then can I assume mass can change its space coordinate with time without any external assistance? Which is undoubtedly not true by a layman thinking. I think I am missing some crucial points, someone, please let me out from this disorder.

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3 Answers 3

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I am applying a force which is perfectly perpendicular to my way to place B

Are you sure about that? Sure, when you aren't moving, the force on the sack is perfectly vertical, but it won't be once you start walking.

Let Y be the vertical axis, and X be the axis from A to B. The force on the sack is not purely in the Y direction, it also has an X component, thus it can do the necessary work.

It may be easier to see what's going on if we use a simple mass on the end of a string, instead of that frictionless sack. As you walk, the string will not remain perfectly vertical, it will point backwards a bit in the X direction.

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  • $\begingroup$ You need to look at the force you're applying to the sack relative to the ground frame. Isn't the force the same in all frames? $\endgroup$ Commented Sep 8, 2019 at 11:26
  • $\begingroup$ It's still somewhat confusing. What is the difference between the force relative to the ground and relative to your body? $\endgroup$ Commented Sep 8, 2019 at 11:57
  • $\begingroup$ @Aaron Well, your body's frame is more complicated because it begins at rest then starts moving. But I'm happy to delete that sentence if you think it will be less confusing without it. $\endgroup$
    – PM 2Ring
    Commented Sep 8, 2019 at 12:11
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If you apply a force to move a mass straight from point A towards point B, then you are applying a force parallel to points A and B. The force will do no perpendicular work, such as moving the mass sideways away from point B.

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If the particle is starting from rest, then you can't apply only a force perpendicular to its motion because there is no motion, and you have to do real work to get the particle started, i.e. the work required to go from 0 to some kinetic energy. You can later do negative work to stop the particle, but the required force will not be strictly perpendicular to the velocity.

If on the other hand the particle starts with some velocity, and you only apply forces that are perpendicular to the motion, then when you finish applying forces, the particle will have the same kinetic energy it started with and indeed, you will have done no work.

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  • $\begingroup$ you can apply work perpendicular to an OBJECT's motion, this concept means that a force in one direction adds no work perpendicular to the FORCE's direction. $\endgroup$ Commented Sep 8, 2019 at 8:26

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