Irodov 1.122: Matching Speeds Link Slope and Floor – Amit & Deepali Sinha Foundation

Problem Irodov 1.122

A disc of mass $m=50$ g starts from rest at the top of an inclined plane that makes an angle $\alpha=30^\circ$ with the horizontal, and slides down it onto a connecting horizontal plane. After covering a distance $l=50$ cm along the horizontal plane, the disc comes to rest. The coefficient of friction $k=0.15$ is the same on both the incline and the horizontal plane. Find the total work performed by the friction forces over the disc’s entire path, from release to rest.

The disc on the incline of angle α, released from rest, with the normal N, the weight m g and the kinetic friction f up the slope. The velocity v points down the slope, toward its foot. — The disc on the incline of angle $\alpha$ , released from rest, with the normal $\vect N$ , the weight $m\vect g$ and the kinetic friction $\vect f$ up the slope. The velocity $\vect v$ points down the slope, toward its foot.

Given

$m=50$ g
$\alpha=30^\circ$
$l=50$ cm
$k=0.15$

Find

A_{\mathrm{fric}}

over the whole path

Solution

Step 1Energy balance on the incline

Let $s$ be the (unknown) length of the incline and $v_1$ the disc’s speed at its foot. Released from rest, the disc gains kinetic energy from gravity’s work $mgs\sin\alpha$ and loses it to friction’s work $-kmg\cos\alpha\, s$ over the slope.

\frac12mv_1^2 = mgs\sin\alpha-kmg\cos\alpha\,s

\frac12mv_1^2 = mgs(\sin\alpha-k\cos\alpha)

(1)

The disc on the horizontal plane just past the foot of the incline, arriving with the speed v_1 and decelerating under friction f alone until it stops after the distance l. — The disc on the horizontal plane just past the foot of the incline, arriving with the speed $v_1$ and decelerating under friction $\vect f$ alone until it stops after the distance $l$ .

Step 2Energy balance on the horizontal plane

On the level plane only friction acts along the motion; the disc’s kinetic energy at the foot of the incline is entirely spent bringing it to rest over the distance $l$ .

0-\frac12mv_1^2 = -kmgl

\frac12mv_1^2 = kmgl

(2)

Step 3Eliminating the incline length

Equating the two expressions for $\frac12mv_1^2$ in (1) and (2) and canceling $mg$ solves for the incline’s own height $h=s\sin\alpha$ , which is all that is needed next.

\begin{align*}s(\sin\alpha-k\cos\alpha) &= kl \\ s\sin\alpha &= kl+ks\cos\alpha\end{align*}

Writing $h=s\sin\alpha$ and $s\cos\alpha=h\cot\alpha$ turns the last line into a single equation for $h$ .

\begin{align*}h &= kl+kh\cot\alpha \\ h(1-k\cot\alpha) &= kl\end{align*}

h = \dfrac{kl}{1-k\cot\alpha}

(3)

Step 4Total work of friction

The disc starts and ends at rest, so gravity’s total work over the whole path, $mgh$ (gravity does no work on the level stretch), is entirely canceled by friction’s total work.

mgh+A_{\mathrm{fric}} = 0

Rearranging the balance leaves $A_{\mathrm{fric}}$ in terms of $h$ alone; substituting (3) for $h$ and the given numbers then gives the algebraic form and its evaluation together.

A_{\mathrm{fric}} = -mgh

A_{\mathrm{fric}} = -\dfrac{kmgl}{1-k\cot\alpha} = -0.05\text{ J}

(4)

Check

Dimensionally $[k][m][g][l]=\mathrm{kg\cdot(m/s^2)\cdot m}=\mathrm J$ and $1-k\cot\alpha$ is dimensionless, so the ratio is an energy, as required. Numerically, $\cot30^\circ=\sqrt3\approx1.73$ , so $1-k\cot\alpha\approx1-0.15(1.73)\approx0.740$ , and $kmgl=0.15(0.050)(9.8)(0.50)\approx0.0368$ J, giving $A_{\mathrm{fric}}\approx-0.0368/0.740\approx-0.050$ J, matching the boxed value.

Remark

The incline’s own length $s$ never had to be found: only its height $h$ survived into the answer, because friction’s work on the slope depends on $s\cos\alpha$ while gravity’s depends on $s\sin\alpha=h$ , and Step 3 combined the two ratios into one equation for $h$ alone. Problem 1.121’s horizontal-projection reading of friction applies here just as it does there: $kmg\cos\alpha\,s$ is $kmg$ times the slope’s horizontal run $s\cos\alpha$ , so friction over the whole path costs $kmg$ times the total horizontal distance covered, the slope’s projection and the level stretch alike. What differs is only what the problem hands the solver. There the base length is given outright, so the answer carries no angle at all; here the slope’s horizontal run is the one length nobody supplies, and it has to be traded for the height through the given $l$ and $k$ . The $\cot\alpha$ in the boxed answer is the exchange rate that trade leaves behind, not the absence of the projection collapse.