# 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? May 14, 2015 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.
– CLR
May 14, 2015 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)? May 14, 2015 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!
– CLR
May 14, 2015 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) May 14, 2015 at 23:27

• 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. Jun 22, 2018 at 2:38