Irodov 1.164: Heat That Doesn't Care Who's Watching – Amit & Deepali Sinha Foundation

Problem Irodov 1.164

A small disc of mass $m$ , released from rest at the top of a smooth hill of height $h$ , slides down the hill and passes directly onto a plank of mass $M$ resting on the horizontal plane at the hill’s base. Friction acts between the disc and the plank (the plank itself slides without friction on the plane beneath it), gradually slowing the disc until, from a certain moment on, the disc and the plank move together as one.

Find the total work done by the friction forces over this whole process.
Determine whether this value is the same in every inertial reference frame.

The disc m slides down the smooth hill, whose foot runs out level with the top of the plank M, and hands off onto it; h is the drop to that level. Friction acts only between disc and plank. — The disc $m$ slides down the smooth hill, whose foot runs out level with the top of the plank $M$ , and hands off onto it; $h$ is the drop to that level. Friction acts only between disc and plank.

Given

Find

(a) the total work

A

done by friction; (b) is

A

frame independent

Solution

Step 1Speed at the foot of the hill

The hill itself is smooth and fixed, so the disc’s descent conserves its own mechanical energy:

\frac{1}{2}mv^2 = mgh \quad\Longrightarrow\quad v=\sqrt{2gh}.

This is the speed the disc carries onto the plank.

Step 2Momentum conservation while friction acts

The floor under the plank is frictionless, so once the disc is on the plank the disc plus plank system feels no external horizontal force: their combined horizontal momentum is conserved from the hand off until they move together at a common velocity $V$ .

mv = (m+M)V.

V = \dfrac{m}{m+M}v

(1)

Step 3Work energy accounting for the friction

Gravity and the normal forces do no work in this purely horizontal sliding, so the only thing that can change the system’s kinetic energy between the hand off and the moment of common velocity is the friction pair itself. The total work $A$ it performs (on the disc and on the plank together) therefore equals the system’s kinetic energy change:

A = \frac{1}{2}(m+M)V^2-\frac{1}{2}mv^2.

Substituting the boxed $V$ from Step 2 and then $v^2=2gh$ from Step 1,

\begin{align*}A &= \frac{1}{2}(m+M)\left(\frac{m}{m+M}v\right)^2-\frac{1}{2}mv^2 = \frac{1}{2}\frac{m^2v^2}{m+M}-\frac{1}{2}mv^2 \\ &= \frac{1}{2}mv^2\left(\frac{m}{m+M}-1\right) = -\frac{1}{2}\frac{mM}{m+M}v^2 = -\frac{mM}{m+M}gh.\end{align*}

A = -\mu gh,\quad \mu=\dfrac{mM}{m+M}

(2)

Step 4Whether the result depends on the reference frame

The disc and the plank feel equal and opposite friction forces at every instant, so together they do work only through their relative sliding, never through either body’s own speed over the ground.

Writing $f$ for the friction force and $v_{\mathrm{rel}}$ for the relative sliding speed between disc and plank at any instant, the total work friction does on the pair is

A = -\int f\,\bigl|v_{\mathrm{rel}}\bigr|\,dt = -\int f\,ds_{\mathrm{rel}},

where $s_{\mathrm{rel}}$ is the relative sliding distance actually traveled, not the distance either body covers over the ground; since $v_{\mathrm{rel}}$ is unchanged by adding a common velocity to both bodies, this integral, and hence the numeric $A$ found above, is the same in every inertial frame, even though the kinetic energies of the disc and the plank separately are not.

Yes,

A

is the same in every inertial frame.

Check

As $M\to\infty$ the plank cannot be moved and $\mu\to m$ , so $A\to -mgh$ : the disc’s entire kinetic energy $mgh$ is dissipated, exactly as it would be sliding to a stop against an immovable floor.

Remark

The same argument holds for any internal dissipative interaction, not just Coulomb friction between rigid surfaces: whenever the interaction force between two bodies does work only through their relative motion, such as viscous drag between them or the contact force in any inelastic collision, the total energy it dissipates is frame independent, while the kinetic energy of either body alone is not, since kinetic energy itself depends on the observer’s frame.