Irodov 1.62: Why Going Up Takes Less Time – Amit & Deepali Sinha Foundation

Problem Irodov 1.62

A small block is launched up a rough incline set at angle $\alpha = 15^\circ$ to the horizontal, decelerates to a stop, and then slides back down under gravity and friction. The time the block takes to climb to its highest point is $\eta = 2.0$ times shorter than the time it takes to slide back down to the starting point. Find the coefficient of kinetic friction between the block and the incline.

The block of mass m launched up the rough incline with speed v_0; the plane stands at angle α to the horizontal and the block returns to the same starting point. — The block of mass $m$ launched up the rough incline with speed $v_0$ ; the plane stands at angle $\alpha$ to the horizontal and the block returns to the same starting point.

Given

$\alpha = 15^\circ$
$\eta = 2.0$

Find

Find the coefficient of friction

k

between the body and the incline

Solution

Step 1The two accelerations, up and down

The forces during the ascent, as this step settles them: the normal N and the weight m g at the center of mass, and kinetic friction f down the slope, opposing the upward motion. On the descent the friction arrow alone reverses. — The forces during the ascent, as this step settles them: the normal $\vect N$ and the weight $m\vect g$ at the center of mass, and kinetic friction $\vect f$ down the slope, opposing the upward motion. On the descent the friction arrow alone reverses.

While the body climbs, gravity’s component along the slope and kinetic friction both resist the upward motion, so both point down the slope and add together.

Newton’s second law along the incline gives the two accelerations,

\begin{aligned} a_1 &= g\sin\alpha + kg\cos\alpha, \\ a_2 &= g\sin\alpha - kg\cos\alpha, \end{aligned}

where $a_1$ is the magnitude of the deceleration on the way up and $a_2$ is the magnitude of the acceleration on the way down.

Step 2The time ratio from kinematics

Let $v_0$ be the launch speed and $d$ the distance reached up the slope. On the ascent the body decelerates from $v_0$ to rest,

0 = v_0 - a_1 t_1, \qquad d = \frac{v_0^2}{2a_1},

which gives $t_1 = v_0/a_1$ . On the descent the body starts from rest and covers the same distance $d$ under $a_2$ ,

d = \frac{1}{2}a_2 t_2^2, \qquad t_2 = \sqrt{\frac{2d}{a_2}} = \sqrt{\frac{v_0^2}{a_1 a_2}} = \frac{v_0}{\sqrt{a_1 a_2}}.

Dividing the two times,

\frac{t_2}{t_1} = \frac{v_0/\sqrt{a_1 a_2}}{v_0/a_1} = \sqrt{\frac{a_1}{a_2}}.

The ascent takes $\eta$ times less than the descent, so $t_1 = t_2/\eta$ , that is $\eta = t_2/t_1$ , and squaring the ratio above gives

\eta^2 = \frac{a_1}{a_2} = \frac{\sin\alpha + k\cos\alpha}{\sin\alpha - k\cos\alpha}

(1)

Step 3Solving for the coefficient of friction symbolically

Cross multiplying (1) and collecting the terms in $k$ on one side isolates it.

\begin{aligned} \eta^2(\sin\alpha - k\cos\alpha) &= \sin\alpha + k\cos\alpha, \\ \sin\alpha(\eta^2 - 1) &= k\cos\alpha(\eta^2+1). \end{aligned}

Dividing by $\cos\alpha(\eta^2+1)$ gives $k$ in closed form,

k = \frac{\eta^2-1}{\eta^2+1}\tan\alpha.

Step 4Evaluating the coefficient

Substituting $\alpha = 15^\circ$ ( $\tan\alpha = 0.268$ ) and $\eta = 2.0$ into the closed form above,

k = \frac{4-1}{4+1}\times 0.268 = 0.6\times 0.268 = 0.161.

k = \frac{\eta^2-1}{\eta^2+1}\tan\alpha = 0.16

(2)

Check

As $\eta \to 1$ (equal ascent and descent times) the formula gives $k \to 0$ : with no friction the two legs are mirror images of each other and take the same time, exactly as expected. The computed value, $k = 0.16$ , is also comfortably below $\tan 15^\circ = 0.27$ , confirming the Step 1 requirement $\tan\alpha > k$ that lets the body slide back down at all.

Remark

The formula gives a way to measure a friction coefficient with nothing but a stopwatch and a protractor: the launch speed $v_0$ never appears in it, having canceled out of the time ratio $t_2/t_1$ in Step 2, so neither the launch speed nor the distance traveled needs to be measured at all.