= Solution
For constant <force> $\mathbf F$ and <displacement> $\mathbf s$, the <work> is
$$
\boxed{W=\mathbf F\cdot\mathbf s=Fs\cos\theta.}
$$
For 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 $\mathbf F\cdot d\mathbf r=m\mathbf a\cdot\mathbf v\,dt=d(mv^2/2)$. 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. \b[Zero net gain of height is compatible with positive muscular work and energy expenditure.]
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