Find the law of motion of a body sliding down a rough…

Find the law of motion of a body sliding down a rough inclined plane, and the time required for it to slide down the entire length, given that it starts from rest at the top. Length $l = 10\text{ m}$, angle of inclination $\alpha = 45^\circ$, coefficient of friction $k = 0.5$, $g = 9.81\text{ m/s}^2$.

Step 1: Forces along the plane

Let $x$ be the distance travelled down the plane from the starting point. Resolving forces along the incline, the component of gravity driving the motion downward is $mg\sin\alpha$; the friction force, proportional to the normal reaction $N=mg\cos\alpha$, opposes the motion:

\[ F_{\parallel} = mg\sin\alpha, \qquad F_f = kN = k\,mg\cos\alpha. \]

The net force down the plane is $F = mg\sin\alpha - k\,mg\cos\alpha = mg(\sin\alpha-k\cos\alpha)$.

Step 2: Equation of motion

By Newton's second law, $m\ddot{x}=F$, so the mass cancels and the acceleration is constant: \[\ddot{x} = g(\sin\alpha - k\cos\alpha) \equiv a. \tag{1}\]

With $\alpha=45^\circ$ (so $\sin\alpha=\cos\alpha=\tfrac{\sqrt2}{2}$) and $k=0.5$: \[a = g\cdot\frac{\sqrt2}{2}(1-0.5) = \frac{g\sqrt2}{4} = \frac{9.81\times1.41421}{4} \approx 3.468\ \text{m/s}^2.\]

Step 3: Integrating the constant-acceleration equation

With initial conditions $x(0)=0$, $\dot{x}(0)=0$, two integrations of (1) give: $\dot{x}=at, \qquad x(t) = \frac{a}{2}\,t^2.$ $x(t) = \frac{g\sqrt2$.

Step 4: Time to slide the entire length

Setting $x(t)=l$ and solving for $t$: \[t = \sqrt{\frac{2l}{a}} = \sqrt{\frac{8l}{g\sqrt2}} = 2\sqrt{\frac{l\sqrt2}{g}}.\] $t = 2\sqrt{\dfrac{l\sqrt2$.

Step 5: Speed at the bottom

From $v=at$, or equivalently from the work–energy relation $v=\sqrt{2al}$: \[v = \sqrt{2al} = \sqrt{\frac{g l\sqrt2}{1}\cdot\frac12}=\sqrt{\frac{gl\sqrt2}{2}} = \sqrt{2\times3.468\times10} \approx 8.33\ \text{m/s}.\]

Step 6: Verification

Two independent routes to $v$ must agree. Using $t\approx2.401\text{ s}$ directly in $v=at$: $v = at \approx 3.468\times2.401 \approx 8.329\ \text{m/s}$.

which matches $v=\sqrt{2al}\approx8.329\ \text{m/s}$ from the energy relation to three figures. $\checkmark$ Dimensionally, $[a]=\text{m/s}^2$, $[t]=\text{s}$ give $[x]=\text{m}$, as required, and setting $k=0$ (frictionless) recovers the familiar $a=g\sin\alpha$.