Irodov 1.134: Friction Doubles Up Against Gravity Uphill

Problem Irodov 1.134

A body of mass $m$ is given an initial speed $v_0$ directed up an inclined plane that makes an angle $\alpha$ with the horizontal, and then released to move on its own. The coefficient of kinetic friction between the body and the incline is $k$ . Find the distance the body travels up the incline before coming to rest, and the work done by friction over that distance.

The body on the incline, moving up-slope with initial speed v_0; weight m g, normal N and kinetic friction f act at its center, friction pointing down-slope since it opposes the upward motion. — The body on the incline, moving up-slope with initial speed $\vect v_0$ ; weight $m\vect g$ , normal $\vect N$ and kinetic friction $\vect f$ act at its center, friction pointing down-slope since it opposes the upward motion.

Given

$m$
$v_0$
$\alpha$
$k$

Find

(a)

s

, the distance traveled before stopping; (b)

A

, the work done by friction over that distance

Solution

Step 1Forces along the incline while moving up

While the body slides up, the incline pushes back with a normal force that balances gravity’s perpendicular component, and along the slope both gravity’s component and kinetic friction act down-slope, opposing the motion.

\begin{align*}N &= mg\cos\alpha \\ ma &= -mg\sin\alpha-kN\end{align*}

a = -g(\sin\alpha+k\cos\alpha)

(1)

Step 2Distance covered before stopping

With the constant deceleration of Step 1, (1), the standard relation between initial speed, final speed and distance gives the stopping distance $s$ once the final speed is set to zero and the result is solved for $s$ .

\begin{align*}0 &= v_0^2+2as \\ s &= -\frac{v_0^2}{2a}\end{align*}

Substituting (1),

s = \dfrac{v_0^2}{2g(\sin\alpha+k\cos\alpha)}

(2)

Step 3Work done by friction over that distance

Friction has constant magnitude $kmg\cos\alpha$ and acts opposite to the body’s displacement throughout, so its work is simply $-kmg\cos\alpha$ times the distance found in Step 2.

\begin{align*}A &= -kmg\cos\alpha\cdot s \\ &= -kmg\cos\alpha\cdot\frac{v_0^2}{2g(\sin\alpha+k\cos\alpha)}\end{align*}

Dividing numerator and denominator by $\cos\alpha$ ,

A = -\dfrac{kmv_0^2}{2(k+\tan\alpha)}

(3)

Check

Dimensionally, $[v_0]^2/[g] = (\mathrm{m/s})^2/(\mathrm{m/s^2}) = \mathrm m$ , so $s$ is a length as required, and $A=-kmg\cos\alpha\cdot s$ then carries units of (force)(length), a joule, as required. As $k\to0$ , $A\to0$ : with no friction the body still stops (gravity alone decelerates it), but friction then does no work at all, exactly as the boxed formula shows.

Remark

Kinetic friction has the same magnitude $kmg\cos\alpha$ whichever way the body moves, so it removes the same amount of mechanical energy, $kmg\cos\alpha\cdot s$ , on a return trip down the same distance $s$ as it removed on the way up. If $\tan\alpha>k$ the body does slide back, and comparing the two legs’ accelerations, $g(\sin\alpha+k\cos\alpha)$ climbing against $g(\sin\alpha-k\cos\alpha)$ descending, shows the descent takes strictly longer than the climb over the same distance. Since friction charges its toll twice, once on the way up and again on the way down, the body returns to the base with kinetic energy $mv_0^2/2-2kmg\cos\alpha\cdot s$ , strictly less than it started with, even though a frictionless incline would return it to the base with exactly $v_0$ by energy conservation.