# A captious work problem: same paths but same forces? [closed]

A man jumps onto a chair. A man climbs onto a chair by putting a leg first and then the other.

In both cases, the work has been the same. TRUE or FALSE...?

Spoiler!: The path is the same, so the change in potential energy is the same. But Work equals potential energy only if there are not non-conservative forces and how can you tell if there are not? Or more simple, how do you know that the change in kinetic energy is the same in both cases?

• Think about what non-conservative forces might act on the man in each case. Are these forces likely to be at all significant compared to the work done against gravity? – tok3rat0r May 14 '15 at 22:25
• I could manage to finish with almost exactly the same velocity in the second case and that would give me 0 work. But if I jump, the change in kinetic energy would be more significant. That's what I'm thinking. – cla May 14 '15 at 22:35
• You said yourself: "The path is the same, so the change in potential energy is the same. But work equals potential energy [if there are no non-conservative forces]". So does the change in kinetic energy matter? In both cases, the man starts with zero velocity on the ground and ends with zero velocity on the chair. Does the change in speed used to get there make any difference (if we can neglect non-conservative forces like friction and air resistance)? – tok3rat0r May 14 '15 at 22:43
• I was thinking that when you jump you land on the chair with a non zero velocity. But maybe considering that was wrong. And my problem was not friction and air resistance, but the forces the man does, wich are completly dissipative. Maybe the aim of the problem was just about the path independence of weight force as I first thought, but I wanted to make sure it wasn't captious. Thanks for the discution! – cla May 14 '15 at 23:07
• (Of course, when jumping onto the chair the man first has to bend his legs to store energy in his muscles ready for the jump. This energy is turned into kinetic energy when he jumps, which is turned into potential energy as he rises to the level of the chair. One could argue that because the process of converting chemical energy into kinetic energy in the muscles is not 100% efficient, he uses slightly more energy when he jumps. But this would be a pretty small effect by any reasonable standard, and in any case it adds nothing to the 'mechanical work' as the term is properly defined) – tok3rat0r May 14 '15 at 23:27

## 3 Answers

The answer depends on how particular you want to be about it. Upon first examination you might say that in the Earth's conservative gravitational field, that the work must be the same, since the start and endpoints are the same and we have ignored any non-conservative forces (e.g. friction, air resistance, etc.). This is more than likely the response that would be expected on an assignment in most classes. One could argue however that if air resistances are taken into account, that more work is done in the jumping case, since one has to combat the higher resistive forces of the air.

I'll answer the question for completeness. As discussed in the comments, if you think about all the posible ways of dissipating energy, maybe you will doubt of the veracity of the assertion made in the exercice. But, the clue is not to think in if the teacher wanted to be "captious"... but that in this case all these ways of dissipating are really small effects and means nothing in comparison to the mechanical energy, and so it is irrelevant for the purpose of the exercice.

Work is the change in kinetic energy. In both cases, the man starts on the ground at rest and end on the chair at rest. In both cases, the net work is zero.

If you want to be more specific and describe the work due to his muscles, that positive work must exactly offset the negative work due to other sources. Gravity does negative work equal to minus the change in gravitational potential energy of the man-earth system, so it doesn't depend on the path. Air drag does some tiny amount of negative work which does depend on the path. That's probably pretty small either way, so it's probably okay to ignore it.

• Work $W=mgh$ in a gravitational field, and there is only Potential Energy stored, no Kinetic Energy, at the end of doing work against a conservative force. The net work done by the man is positive. A gravitational field is a conservative force, and work done against a gravitational field alone leaves no remnant of Kinetic Energy. The energy expended to move his muscles will at least equal the Potential Energy stored in the Gravitational Field. And yes, there will be non-conservative forces acting within the muscles, and on air, which will produce random Kinetic Energy/Thermal Energy. – Thomas Lee Abshier ND Jun 22 '18 at 2:38