Try it yourself on a set of wide steps! Work is given by
$$\int_C \mathbf{F} \cdot d\mathbf{x}$$ where $C$ is a path integral. In this case I think $\mathbf{F}$ is a rotational vector field because the stairs are essentially a set of discontinuities. This would mean that the integral is path dependent.
Here is why this seems like a paradox: if you where to climb the same height over a smooth hill, then the integral is path independent since $\mathbf{F}$ is "smooth" and irrotational. This means that the amount of work done never changes given the path you take, hence if you take a longer (diagonal) path you must exert less force over that distance than if you where to take a straight path. In the case of walking up stairs, you must always cross the same number of stairs no matter the path you take. All other distance you make is along the horizontal parts of the tops of stairs and hence requires much less force. This would seem to suggest to me walking up stairs always requires the same force to scale the vertical parts of each stair.
Am I hallucinating that walking diagonally up stairs is easier?
Best Answer
Ascending a hill diagonally means that while the work is the same, the power expenditure is lower (but it takes longer), which is why it is easier.