 Solution

For constant force $\mathbf{F}$ and displacement $\mathbf{s}$, the work isFor a variable force, $W={\int }_C\mathbf{F}\cdot d\mathbf{r}$. The work-energy theorem explains this choice: Newton's second law for fixed mass gives $\mathbf{F}\cdot d\mathbf{r}=m\mathbf{a}\cdot \mathbf{v}dt=d(m{v}^2/2)$. The component along the motion changes kinetic energy; a perpendicular force can change direction without doing work. Exerting a force on a stationary object likewise does no work on it.

Walking on level ground still requires muscular work. The centre of mass rises and falls during each step; the limbs gain and lose kinetic energy; and muscles perform positive work that replaces energy dissipated in impacts and tissue motion. Air resistance can add another cost. Returning to the same height makes the net change of gravitational potential energy zero, not every intermediate work contribution.

At ideal no-slip contact with an immovable road, the contact point is instantaneously stationary, so the road's force does no work through that point. The muscles convert internal chemical energy into motion. Zero net gain of height is compatible with positive muscular work and energy expenditure.