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PhysicsLaws of Motion
A block slides down a smooth incline in time $t$. The same block slides down an identical rough inclined plane in time $2t$. The coefficient of kinetic friction is (angle of inclination $= 45°$):
Options
1
$0.25$
2
$0.50$
3
$0.75$
4
$1.00$
Correct Answer
$\mu_k = 0.75$
Solution
1

Smooth incline: $a_1 = g\sin 45° = \dfrac{g}{\sqrt{2}}$

Rough incline: $a_2 = g(\sin 45° - \mu_k\cos 45°) = \dfrac{g(1-\mu_k)}{\sqrt{2}}$

2

Same distance, $t_{smooth}=t$, $t_{rough}=2t$:

$$\frac{1}{2}a_1 t^2 = \frac{1}{2}a_2(2t)^2 \implies a_1 = 4a_2$$$$\frac{g}{\sqrt{2}} = \frac{4g(1-\mu_k)}{\sqrt{2}} \implies 1 = 4(1-\mu_k) \implies \boxed{\mu_k = 0.75}$$
$a_1 = 4a_2 \Rightarrow 1 = 4(1-\mu_k) \Rightarrow \mu_k = 0.75$
Theory: Laws of Motion
1. Motion on Inclined Plane

On smooth incline (angle $\theta$): $a = g\sin\theta$. On rough incline: $a = g(\sin\theta - \mu_k\cos\theta)$. Normal force $N = mg\cos\theta$. Friction $f = \mu_k N = \mu_k mg\cos\theta$. Block just slides if $\tan\theta > \mu_s$ (angle of friction). Angle of repose = $\tan^{-1}(\mu_s)$.

2. Newton's Laws

$\vec{F}_{net} = m\vec{a}$. Free body diagram isolates one object and shows all forces ON it. For connected bodies: write equations for each body separately. Constraint: if connected by string over pulley, magnitudes of acceleration are equal. Atwood machine: $a = (m_1-m_2)g/(m_1+m_2)$. Tension $T = 2m_1m_2g/(m_1+m_2)$.

3. Friction

Static friction $f_s \leq \mu_s N$ (self-adjusting, up to maximum). Kinetic friction $f_k = \mu_k N$ (constant once sliding). $\mu_k < \mu_s$ (kinetic always less than static). Rolling friction $\ll$ sliding friction (why wheels used). Friction is independent of contact area (Amontons law). Depends only on nature of surfaces and normal force.

4. Work-Energy Theorem

$W_{net} = \Delta KE = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$. Work by friction $= -\mu_k mgs\cos\theta$ (always negative, reduces KE). For block on rough incline: $\Delta KE = mgs\sin\theta - \mu_k mgs\cos\theta$. Energy dissipated by friction $= \mu_k Ns$ = heat generated.

5. Circular Motion on Roads

Flat road: friction provides centripetal force. Max safe speed: $v_{max} = \sqrt{\mu_s Rg}$. Banked road: normal force has horizontal component for centripetal. Optimal banking angle: $\tan\theta = v^2/(Rg)$ (no friction needed). With friction: $v_{max} = \sqrt{Rg(\tan\theta+\mu_s)/(1-\mu_s\tan\theta)}$.

6. Impulse and Momentum

Impulse $J = F\Delta t = \Delta p = mv - mu$. Conservation of momentum: when $F_{ext} = 0$, total momentum conserved. Elastic collision: KE conserved. $e = 1$ (coefficient of restitution). Inelastic: KE not conserved. $e < 1$. Perfectly inelastic: bodies stick together. $e = 0$. For head-on elastic collision: $v_1 = (m_1-m_2)u_1/(m_1+m_2)$, $v_2 = 2m_1u_1/(m_1+m_2)$.

7. Rotational Dynamics

$\tau = I\alpha$ (rotational analogue of $F = ma$). Moment of inertia $I = \sum m_i r_i^2$. Key values: disc $I = MR^2/2$, ring $I = MR^2$, solid sphere $I = 2MR^2/5$, hollow sphere $I = 2MR^2/3$, rod (centre) $I = ML^2/12$. Parallel axis theorem: $I = I_{cm} + Md^2$. Perpendicular axis theorem (lamina): $I_z = I_x + I_y$.

8. Rolling Motion

Rolling without slipping: $v_{cm} = R\omega$. Total KE $= \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv^2(1 + I/mR^2)$. For disc: factor $= 1 + 1/2 = 3/2$. For ring: $1 + 1 = 2$. For sphere: $1 + 2/5 = 7/5$. Sphere rolls fastest down incline (smallest $I/mR^2$). Ring rolls slowest (largest $I/mR^2 = 1$).

Frequently Asked Questions
1. Why is $a_1 = 4a_2$? ⌄
$s = \frac{1}{2}a_1t^2$ (smooth). Same distance $s = \frac{1}{2}a_2(2t)^2 = 2a_2t^2$ (rough). So $\frac{1}{2}a_1t^2 = 2a_2t^2 \Rightarrow a_1 = 4a_2$. The time is doubled on rough surface, so acceleration must be 4 times smaller (since $s \propto at^2$, for same $s$: $a \propto 1/t^2$, and $(2t)^2 = 4t^2$).
2. What if angle were 30° instead of 45°? ⌄
With $\theta = 30°$: $a_1 = g\sin30° = g/2$. $a_2 = g(\sin30° - \mu_k\cos30°) = g(1/2 - \mu_k\sqrt{3}/2)$. From $a_1 = 4a_2$: $1/2 = 4(1/2 - \mu_k\sqrt{3}/2) = 2 - 2\sqrt{3}\mu_k$. $2\sqrt{3}\mu_k = 3/2$. $\mu_k = 3/(4\sqrt{3}) = \sqrt{3}/4 \approx 0.43$.
3. What is angle of repose? ⌄
The angle of repose is the maximum angle of inclination at which a block remains stationary on a rough incline. At angle of repose $\theta_r$: $mg\sin\theta_r = \mu_s mg\cos\theta_r$, so $\tan\theta_r = \mu_s$. If the incline is steeper than $\theta_r$, the block slides. For $\mu_k = 0.75$: angle of repose $= \tan^{-1}(0.75) \approx 36.9°$.
4. Can μk ever be greater than 1? ⌄
Yes. Friction coefficient can be greater than 1 for certain material pairs. $\mu_k = 0.75$ in this problem. Rubber on dry concrete: $\mu_s \approx 0.7$–$1.0$. Gecko feet: $\mu$ effectively very high (van der Waals forces). Silicon: up to 1.5. No physical law limits $\mu \leq 1$; that is a common misconception. However, most engineering materials have $\mu < 1$.
5. What is the energy dissipated by friction in this problem? ⌄
Energy dissipated $= f_k \times s = \mu_k mg\cos\theta \times s$. We know $a_2 = g/\sqrt{2} \times 1/4$ (since $a_1 = 4a_2$ and $a_1 = g/\sqrt{2}$). Distance $s = \frac{1}{2}a_2(2t)^2 = 2a_2t^2 = 2 \times \frac{g}{4\sqrt{2}} \times t^2$. Energy dissipated $= 0.75 \times mg\cos45° \times s = 0.75 \times mg/\sqrt{2} \times s$. (Numerical value needs $m$ and $t$, but the formula shows it depends on $\mu_k$, $m$, $g$, $\theta$, and time.)
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