Irodov 1.121: Only Height and Base Length Matter – Amit & Deepali Sinha Foundation

Problem Irodov 1.121

A body of mass $m$ is hauled slowly up a hill of arbitrary profile, height $h$ and base length $l$ , by a force $\vect F$ that, at every point of the path, is directed tangent to the trajectory. The coefficient of friction between the body and the hill’s surface is $k$ . Find the work done by the force $\vect F$ over the whole ascent.

The hill’s profile, rising a height $h$ over a horizontal base of length $l$ . The body of mass $m$ is hauled up by the force $\vect F$ , kept tangent to the curve at every point.

Given

Find

A_F

Solution

Step 1Zero net kinetic energy change over the slow ascent

“Slowly” means the body’s speed stays negligible throughout, so its kinetic energy is unchanged from start to finish. The work-energy theorem then sets the sum of every force’s work to zero.

A_F + A_{\mathrm{grav}} + A_{\mathrm{fric}} = 0 \tag{1}

Step 2Work of gravity

Gravity does work only through the net rise $h$ , regardless of the path taken to gain it: integrating the weight’s downward pull against the element of upward displacement $dy$ over the whole climb gives the total.

A_{\mathrm{grav}} = -\int_0^h mg\,dy

A_{\mathrm{grav}} = -mgh

(2)

Step 3Work of friction from the horizontal projection

At a point where the track makes an angle $\theta$ with the horizontal, the normal force is $N=mg\cos\theta$ and the element of friction work is $-kN\,ds=-kmg\cos\theta\,ds$ . Since $\cos\theta\,ds=dx$ is the element’s own horizontal projection, the integral collapses to the base length $l$ alone.

\begin{align*}A_{\mathrm{fric}} &= -\int kmg\cos\theta\,ds \\ &= -kmg\int_0^l dx\end{align*}

A_{\mathrm{fric}} = -kmgl

(3)

Step 4The work of

\vect F

Substituting (2) and (3) into the balance (1) and solving for $A_F$ gives the answer.

A_F - mgh - kmgl = 0

A_F = mg(h+kl)

(4)

Check

Dimensionally $[m][g][h]=\mathrm{kg}\cdot\mathrm{m/s^2}\cdot\mathrm m=\mathrm J$ , an energy, and $[k][l]$ carries the same units as $h$ since $k$ is dimensionless, so the sum inside the parentheses is consistently a length. With $k=0$ , the hill is frictionless and $A_F\to mgh$ , exactly the work needed to raise the body through $h$ and nothing more, matching a smooth ramp of any shape.

Remark

The result depends on the hill only through $h$ and $l$ , never through its shape or the path length actually traversed: a steep, short climb and a gentle, long one with the same $h$ and $l$ cost the hauling force exactly the same work. That is the content of the horizontal-projection trick in Step 3: friction’s element of work $-kmg\cos\theta\,ds$ collapses to $-kmg\,dx$ , and $dx$ only integrates up to the base length $l$ because the path is monotone, always advancing toward the far end of the base. Gravity’s own path-independence needs no such restriction, since $dy$ integrates to $h$ on any path at all; friction’s version borrows the same trick only for a profile that never doubles back, and even then only because $k$ and $g$ are the same constant at every point of this particular hill.