Motion (or rest) on Rough Inclined Surface Questions in English

(B) Consider a body of mass $m$ on an inclined plane with angle $\theta$. The forces acting on the body are:
$1$. The component of gravitational force acting down the plane: $mg \sin \theta$.
$2$. The normal force perpendicular to the plane: $N = mg \cos \theta$.
$3$. The kinetic frictional force acting up the plane: $f_k = \mu N = \mu mg \cos \theta$.
The net force $F_{\text{net}}$ acting down the plane is the difference between the gravitational component and the frictional force:
$F_{\text{net}} = mg \sin \theta - f_k = mg \sin \theta - \mu mg \cos \theta$.
Using Newton's second law,$F_{\text{net}} = ma$,we get:
$ma = mg \sin \theta - \mu mg \cos \theta$.
Dividing both sides by $m$,the acceleration $a$ is:
$a = g(\sin \theta - \mu \cos \theta)$.

(A) When a body moves up an inclined plane,the forces acting against the motion are the component of gravity along the plane $(mg \sin \theta)$ and the frictional force $(f = \mu N = \mu mg \cos \theta)$.
According to Newton's second law,the net force $F_{net} = ma = -(mg \sin \theta + \mu mg \cos \theta)$.
Thus,the retardation $a$ is given by $a = g(\sin \theta + \mu \cos \theta)$.
Substituting the given values $\theta = 30^{\circ}$ and $\mu = 0.5$:
$a = g(\sin 30^{\circ} + 0.5 \cos 30^{\circ})$
$a = g\left(\frac{1}{2} + 0.5 \times \frac{\sqrt{3}}{2}\right)$
$a = g\left(\frac{1}{2} + \frac{\sqrt{3}}{4}\right)$
$a = g\left(\frac{2 + \sqrt{3}}{4}\right)$.

(A) The block stops on the inclined plane when its initial kinetic energy is used to do work against friction and is partly converted into potential energy.
Given the angle of inclination is $\theta = 45^{\circ}$,the distance covered on the incline $(s)$ is related to the vertical height $(h)$ as:
$\sin 45^{\circ} = \frac{h}{s} \implies s = \frac{h}{\sin 45^{\circ}} = h \sqrt{2}$.
According to the work-energy theorem,the initial kinetic energy $(K.E.)$ is equal to the work done against friction plus the gain in potential energy:
$K.E. = W_{\text{friction}} + \Delta U$
$K.E. = (\mu mg \cos \theta) \cdot s + mgh$
Substituting $s = h \sqrt{2}$ and $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$:
$100 = (0.5 \times 5 \times 10 \times \frac{1}{\sqrt{2}}) \times (h \sqrt{2}) + (5 \times 10 \times h)$
$100 = (0.5 \times 5 \times 10 \times h) + 50h$
$100 = 25h + 50h$
$100 = 75h$
$h = \frac{100}{75} = \frac{4}{3} \text{ m}$.
Now,the distance covered $s$ is:
$s = h \sqrt{2} = \frac{4}{3} \sqrt{2} \text{ m}$.

(A) Consider a body of mass $m$ on an inclined plane with angle $\theta$. The forces acting on the body are:
$1$. Gravitational force $mg$ acting vertically downwards.
$2$. Normal reaction $N$ perpendicular to the inclined plane.
$3$. Frictional force $f$ acting up the plane,opposing the motion.
The components of the gravitational force are $mg \sin \theta$ (parallel to the plane) and $mg \cos \theta$ (perpendicular to the plane).
For equilibrium perpendicular to the plane,$N = mg \cos \theta$.
The frictional force is $f = \mu N = \mu mg \cos \theta$.
Applying Newton's second law along the plane:
$mg \sin \theta - f = ma$
$mg \sin \theta - \mu mg \cos \theta = ma$
Dividing by $m$,we get the acceleration:
$a = g(\sin \theta - \mu \cos \theta)$

(A) When a body of mass $m$ slides down an inclined plane with acceleration $a$,the forces acting on the body are as follows:
$1$. The component of gravitational force acting down the plane is $mg \sin \theta$.
$2$. The normal reaction force $R$ is equal to $mg \cos \theta$.
$3$. The frictional force acting up the plane is $f = \mu R = \mu mg \cos \theta$.
Applying Newton's second law of motion along the plane:
$ma = mg \sin \theta - f$
$ma = mg \sin \theta - \mu mg \cos \theta$
Dividing both sides by $m$:
$a = g \sin \theta - \mu g \cos \theta$
$a = g(\sin \theta - \mu \cos \theta)$

(B) Let the particle be at an angle $\theta$ with the vertical. The forces acting on the particle are its weight $mg$ downwards,the normal force $N$ radially outwards,and the frictional force $f$ tangential to the surface.
For the particle to remain stationary,the component of weight along the tangent must be balanced by the limiting friction: $mg \sin \theta = f = \mu N$.
The component of weight perpendicular to the surface is balanced by the normal force: $N = mg \cos \theta$.
Substituting $N$ into the friction equation: $mg \sin \theta = \mu (mg \cos \theta) \implies \tan \theta = \mu = \frac{1}{\sqrt{3}}$.
Thus,$\theta = 30^\circ$.
The height $h$ of the particle from the bottom of the hemisphere is $h = R - R \cos \theta = R(1 - \cos 30^\circ)$.
$h = R(1 - \frac{\sqrt{3}}{2})$.

(A) Let the total length of the inclined plane be $l$. The upper half has length $l/2$ and is smooth,while the lower half has length $l/2$ and is rough with coefficient of friction $\mu$.
For the upper half (smooth):
The acceleration is $a_1 = g \sin \phi$. The initial velocity $u = 0$. Using the equation $v^2 = u^2 + 2as$,the velocity $v$ at the midpoint is:
$v^2 = 0 + 2(g \sin \phi)(l/2) = gl \sin \phi$.
For the lower half (rough):
The initial velocity is $v$ (from the midpoint),and the final velocity is $0$ (at the bottom). The acceleration is $a_2 = g(\sin \phi - \mu \cos \phi)$. Using $v_f^2 = v_i^2 + 2a_2s$:
$0 = v^2 + 2g(\sin \phi - \mu \cos \phi)(l/2)$.
Substituting $v^2 = gl \sin \phi$:
$0 = gl \sin \phi + gl(\sin \phi - \mu \cos \phi)$.
Dividing by $gl$:
$0 = \sin \phi + \sin \phi - \mu \cos \phi$.
$2 \sin \phi = \mu \cos \phi$.
Therefore,$\mu = 2 \tan \phi$.

(D) For a smooth inclined plane,the acceleration is $a_s = g \sin \theta$. The time taken to cover distance $s$ is $t_s = \sqrt{\frac{2s}{g \sin \theta}}$.
For a rough inclined plane,the acceleration is $a_r = g(\sin \theta - \mu \cos \theta)$. The time taken is $t_r = \sqrt{\frac{2s}{g(\sin \theta - \mu \cos \theta)}}$.
Given $t_r = n t_s$,where $n = 2$.
Squaring both sides: $\frac{2s}{g(\sin \theta - \mu \cos \theta)} = n^2 \frac{2s}{g \sin \theta}$.
This simplifies to $\sin \theta = n^2(\sin \theta - \mu \cos \theta)$.
Rearranging for $\mu$: $\mu = \tan \theta \left(1 - \frac{1}{n^2}\right)$.
Substituting $\theta = 45^{\circ}$ and $n = 2$: $\mu = \tan 45^{\circ} \left(1 - \frac{1}{2^2}\right) = 1 \times (1 - 0.25) = 0.75$.

(C) When a particle is projected up an inclined plane,both the component of gravity acting down the plane and the frictional force act in the direction opposite to the motion.
The net force acting on the particle is $F_{net} = -(mg \sin \theta + f_k)$,where $f_k = \mu_k N = \mu_k mg \cos \theta$.
Using Newton's second law,$ma = -(mg \sin \theta + \mu_k mg \cos \theta)$.
The magnitude of the retardation (deceleration) is $a = g(\sin \theta + \mu \cos \theta)$.
Given $\theta = 45^{\circ}$ and $\mu = 0.5$,we have:
$a = g(\sin 45^{\circ} + 0.5 \cos 45^{\circ})$
$a = g(\frac{1}{\sqrt{2}} + 0.5 \cdot \frac{1}{\sqrt{2}})$
$a = g(\frac{1}{\sqrt{2}} + \frac{1}{2 \sqrt{2}}) = g(\frac{2+1}{2 \sqrt{2}})$
$a = \frac{3g}{2 \sqrt{2}}$.

(D) The maximum possible acceleration of the person on the conveyor belt is given by the net force along the incline divided by mass:
$a_{\max} = \frac{\mu m g \cos \theta - m g \sin \theta}{m} = g(\mu \cos \theta - \sin \theta)$
Given $\theta = 30^{\circ}$,$\mu = \frac{5}{3 \sqrt{3}}$,and $\sin 30^{\circ} = 0.5$,$\cos 30^{\circ} = \frac{\sqrt{3}}{2}$:
$a_{\max} = g \left( \frac{5}{3 \sqrt{3}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{2} \right) = g \left( \frac{5}{6} - \frac{3}{6} \right) = \frac{g}{3}$
The length of the conveyor belt $L$ is related to height $h$ by $L = \frac{h}{\sin 30^{\circ}} = 2h$.
The initial velocity $u$ of the person relative to the ground is the speed of the belt,$u = \sqrt{\frac{g h}{6}}$.
Using the equation of motion $S = ut + \frac{1}{2} a t^2$:
$2h = \left( \sqrt{\frac{g h}{6}} \right) t + \frac{1}{2} \left( \frac{g}{3} \right) t^2$
$2h = \frac{t \sqrt{gh}}{\sqrt{6}} + \frac{gt^2}{6}$
Multiplying by $6$:
$12h = t \sqrt{6gh} + gt^2$
$gt^2 + t \sqrt{6gh} - 12h = 0$
Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$t = \frac{-\sqrt{6gh} + \sqrt{6gh - 4(g)(-12h)}}{2g} = \frac{-\sqrt{6gh} + \sqrt{6gh + 48gh}}{2g} = \frac{-\sqrt{6gh} + \sqrt{54gh}}{2g} = \frac{-\sqrt{6gh} + 3\sqrt{6gh}}{2g} = \frac{2\sqrt{6gh}}{2g} = \sqrt{\frac{6h}{g}}$

(D) Given: Mass $m = 2 \ kg$,acceleration $a = 4 \ m/s^2$,angle $\theta = 30^{\circ}$,$g = 10 \ m/s^2$.
Case $1$: Body sliding down.
The component of gravitational force along the plane is $mg \sin \theta = 2 \times 10 \times \sin 30^{\circ} = 20 \times 0.5 = 10 \ N$.
Let $f$ be the frictional force acting upwards. The equation of motion is:
$mg \sin \theta - f = ma$
$10 - f = 2 \times 4$
$10 - f = 8$
$f = 2 \ N$.
Case $2$: Body moving up with the same acceleration.
Let $F$ be the external force applied upwards. The frictional force $f$ will now act downwards.
The equation of motion is:
$F - mg \sin \theta - f = ma$
$F - 10 - 2 = 2 \times 4$
$F - 12 = 8$
$F = 20 \ N$.

(C) An object begins to slide down an inclined plane when the angle of inclination (also called the angle of repose) is equal to the angle of friction. Let $\theta$ be the angle at which a mass begins to slide down an incline.
At the verge of sliding,the force of static friction $f_s$ is equal to the downward component of gravity.
$f_s = mg \sin \theta$
Since $f_s = \mu N$ and the normal force $N = mg \cos \theta$,we have:
$\mu (mg \cos \theta) = mg \sin \theta$
$\mu = \frac{\sin \theta}{\cos \theta} = \tan \theta$
$\theta = \tan^{-1} \mu$
Since the angle of repose $\theta$ depends only on the coefficient of static friction $\mu$ and is independent of the mass $m$ of the body,all three masses will begin to slide at the same instant at the same inclination angle $\theta$.

(B) $1$. When the object slides down with uniform velocity,the net force is zero. This implies the component of gravity $mg \sin \theta$ is balanced by the kinetic friction $f_k = \mu_k N = \mu_k mg \cos \theta$. Thus,$\mu_k = \tan \theta$.
$2$. When the object is pushed up with velocity $u$,it experiences a retardation $a_{up} = g \sin \theta + \mu_k g \cos \theta$. Since $\mu_k = \tan \theta$,$a_{up} = g \sin \theta + (\tan \theta) g \cos \theta = 2g \sin \theta$.
$3$. After stopping,the object slides down. The acceleration while sliding down is $a_{down} = g \sin \theta - \mu_k g \cos \theta$. Since $\mu_k = \tan \theta$,$a_{down} = g \sin \theta - g \sin \theta = 0$. Wait,this implies uniform velocity. Let's re-evaluate: the kinetic friction acts opposite to motion. When sliding down,$a_{down} = g \sin \theta - \mu_k g \cos \theta = 0$. The object will slide down with the same velocity it had at the start of the descent. Since it started from rest at the top,it will slide down with uniform velocity. However,the question implies it reaches the ground. The energy lost due to friction during the upward journey is $W_f = f_k \times d = (\mu_k mg \cos \theta) d$. During the downward journey,the same energy is lost. By work-energy theorem,the final kinetic energy $K_f = K_i + W_{gravity} - W_{friction}$. Since $W_{gravity}$ is the same for both paths and $W_{friction}$ is the same,the final velocity will be less than $u$ due to energy dissipation.

(D) Let $h$ be the height of the sand cone with base radius $R$. The angle of repose $\alpha$ is the maximum angle at which the sand can be piled without slipping. This angle is related to the coefficient of static friction $\mu$ by $\tan \alpha = \mu$.
In the cone,the angle $\alpha$ is formed by the slant height with the horizontal ground. From the geometry of the cone,$\tan \alpha = \frac{h}{R}$.
Equating the two expressions for $\tan \alpha$,we get $\frac{h}{R} = \mu$,which implies $h = \mu R$.
The volume $V$ of a cone is given by $V = \frac{1}{3} \pi R^2 h$.
Substituting $h = \mu R$ into the volume formula,we get $V_{\max} = \frac{1}{3} \pi R^2 (\mu R) = \frac{\mu \pi R^3}{3}$.

(D) Let $l$ be the total length of the inclined plane and $\theta$ be the angle of inclination.
According to the work-energy theorem,the total work done on the body is equal to the change in its kinetic energy.
Since the body starts from rest and comes to rest at the bottom,the change in kinetic energy $\Delta K = 0$.
The forces acting on the body are gravity,friction,and the normal force.
$1$. Work done by gravity $(W_g)$: $W_g = mgh = mg(l \sin \theta)$.
$2$. Work done by friction $(W_f)$: Friction acts only on the lower part of length $l(1 - 1/n)$. Thus,$W_f = -f_k \cdot d = -(\mu_k mg \cos \theta) \cdot l(1 - 1/n)$.
$3$. Work done by the normal force $(W_N)$: Since the normal force is always perpendicular to the displacement,$W_N = 0$.
Applying the work-energy theorem: $W_g + W_f + W_N = 0$.
$mg l \sin \theta - \mu_k mg \cos \theta \cdot l \left(\frac{n-1}{n}\right) = 0$.
Dividing by $mg l \cos \theta$ (assuming $\cos \theta \neq 0$):
$\tan \theta = \mu_k \left(\frac{n-1}{n}\right)$.
Therefore,$\theta = \tan^{-1}\left[\left(\frac{n-1}{n}\right) \mu_k\right]$.

(B) For a block to remain steady on an accelerating wedge,the forces acting on the block in the frame of the wedge are: gravity ($mg$ downwards),pseudo force ($ma$ horizontally outwards),normal force ($N$ perpendicular to the surface),and friction ($f$ parallel to the surface).
To find the minimum acceleration $(a_{\min})$,friction must act upwards along the incline to prevent the block from sliding down.
Resolving forces parallel and perpendicular to the incline:
$N = mg \cos \theta + ma \sin \theta$
$f + ma \cos \theta = mg \sin \theta$
Since $f = \mu N$,we have $\mu(mg \cos \theta + ma \sin \theta) = mg \sin \theta - ma \cos \theta$.
Rearranging for $a$:
$ma(\mu \sin \theta + \cos \theta) = mg(\sin \theta - \mu \cos \theta)$
$a = g \frac{\sin \theta - \mu \cos \theta}{\cos \theta + \mu \sin \theta}$
Given $\theta = 45^{\circ}$,$\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$.
$a_{\min} = g \frac{\frac{1}{\sqrt{2}} - \mu \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}} + \mu \frac{1}{\sqrt{2}}} = g \frac{1 - \mu}{1 + \mu}$
Substituting $g = 10 \ m \ s^{-2}$ and $\mu = 0.4$:
$a_{\min} = 10 \times \frac{1 - 0.4}{1 + 0.4} = 10 \times \frac{0.6}{1.4} = 10 \times \frac{6}{14} = \frac{60}{14} = \frac{30}{7} \ m \ s^{-2}$.

(B) The frictional force acting on both blocks is kinetic in nature.
For the $1 \ kg$ block:
$f_1 = \mu m_1 g \cos 45^{\circ} = 0.4 \times 1 \times 10 \times \frac{1}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} \ N$
For the $2 \ kg$ block:
$f_2 = \mu m_2 g \cos 45^{\circ} = 0.4 \times 2 \times 10 \times \frac{1}{\sqrt{2}} = \frac{8}{\sqrt{2}} = 4\sqrt{2} \ N$
The net force along the incline in the downward direction provides the acceleration $a = \alpha \sqrt{2}$:
$(m_1 + m_2)g \sin 45^{\circ} - (f_1 + f_2) = (m_1 + m_2)a$
$(1 + 2) \times 10 \times \frac{1}{\sqrt{2}} - (2\sqrt{2} + 4\sqrt{2}) = (1 + 2) \times (\alpha \sqrt{2})$
$30 \times \frac{1}{\sqrt{2}} - 6\sqrt{2} = 3\alpha \sqrt{2}$
Multiply throughout by $\sqrt{2}$:
$30 - 6 \times 2 = 3\alpha \times 2$
$30 - 12 = 6\alpha$
$18 = 6\alpha$
$\alpha = 3$

(B) The block will start slipping when the slope of the curve at that point is equal to the tangent of the angle of repose,$\tan \theta = \mu_s$.
The slope of the curve at any point $x$ is given by the derivative:
$m = \frac{dy}{dx} = \frac{d}{dx} \left( \frac{x^2}{20} \right) = \frac{2x}{20} = \frac{x}{10}$.
For the block to remain in equilibrium without slipping,the slope must satisfy:
$\frac{dy}{dx} \leq \mu_s$
At the point of slipping,$\frac{x}{10} = 0.5$.
Solving for $x$:
$x = 0.5 \times 10 = 5 \ m$.
Now,substitute $x = 5 \ m$ into the equation of the parabola to find the maximum height $h$:
$h = y = \frac{x^2}{20} = \frac{5^2}{20} = \frac{25}{20} = 1.25 \ m$.

(B) Given: Mass $m = 4 \ kg$,coefficient of static friction $\mu_s = 0.5$,and frictional force $f = 14.14 \ N = 10\sqrt{2} \ N$.
Since the block is at rest on the inclined plane,the frictional force balances the component of gravitational force acting down the plane.
$f = mg \sin \theta$
$10\sqrt{2} = 4 \times 10 \times \sin \theta$
$10\sqrt{2} = 40 \sin \theta$
$\sin \theta = \frac{10\sqrt{2}}{40} = \frac{\sqrt{2}}{4} = \frac{1}{2\sqrt{2}} \approx 0.3535$.
However,if we assume the frictional force given is the limiting friction $(f_L = \mu_s N)$:
Normal reaction $N = mg \cos \theta = 40 \cos \theta$.
$f_L = \mu_s N = 0.5 \times 40 \cos \theta = 20 \cos \theta$.
Equating $f_L = 10\sqrt{2}$:
$20 \cos \theta = 10\sqrt{2}$
$\cos \theta = \frac{10\sqrt{2}}{20} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.
Therefore,$\theta = 45^{\circ}$.

(A) To find the maximum angle $\theta$ for which the block remains at rest,consider the forces acting on the block of mass $m$:
$1$. The gravitational force $mg$ acts vertically downwards.
$2$. The normal reaction $N$ acts perpendicular to the inclined surface.
$3$. The static friction force $f_s$ acts up the incline to oppose the tendency of motion.
Resolving the weight $mg$ into components:
- Component perpendicular to the incline: $mg \cos \theta$
- Component parallel to the incline: $mg \sin \theta$
For equilibrium perpendicular to the incline:
$N = mg \cos \theta$
For the block to remain motionless,the driving force must be less than or equal to the limiting friction:
$mg \sin \theta \leq f_{s, \text{max}}$
Since $f_{s, \text{max}} = \mu N = \mu mg \cos \theta$,we have:
$mg \sin \theta \leq \mu mg \cos \theta$
Dividing both sides by $mg \cos \theta$ (assuming $\cos \theta \neq 0$):
$\tan \theta \leq \mu$
Thus,the maximum value of $\theta$ is given by $\tan \theta = \mu$.

(D) Let $a$ be the common acceleration of the two blocks.
For the first block $(m_1)$: The forces acting along the incline are the component of gravity $m_1 g \sin 60^{\circ}$ downwards,friction force $f_1 = \mu_1 N_1 = 1.5 \mu m_1 g \cos 60^{\circ}$ upwards,and the reaction force $R$ from the second block acting upwards.
Equation of motion: $m_1 g \sin 60^{\circ} - 1.5 \mu m_1 g \cos 60^{\circ} + R = m_1 a$ --- $(i)$
For the second block $(m_2)$: The forces acting along the incline are the component of gravity $m_2 g \sin 60^{\circ}$ downwards,friction force $f_2 = \mu_2 N_2 = \mu m_2 g \cos 60^{\circ}$ upwards,and the reaction force $R$ from the first block acting downwards.
Equation of motion: $m_2 g \sin 60^{\circ} - \mu m_2 g \cos 60^{\circ} - R = m_2 a$ --- $(ii)$
From $(i)$,$a = g \sin 60^{\circ} - 1.5 \mu g \cos 60^{\circ} + \frac{R}{m_1}$.
From $(ii)$,$a = g \sin 60^{\circ} - \mu g \cos 60^{\circ} - \frac{R}{m_2}$.
Equating the two expressions for $a$:
$g \sin 60^{\circ} - 1.5 \mu g \cos 60^{\circ} + \frac{R}{m_1} = g \sin 60^{\circ} - \mu g \cos 60^{\circ} - \frac{R}{m_2}$
$R \left( \frac{1}{m_1} + \frac{1}{m_2} \right) = 1.5 \mu g \cos 60^{\circ} - \mu g \cos 60^{\circ} = 0.5 \mu g \cos 60^{\circ}$
$R \left( \frac{m_1 + m_2}{m_1 m_2} \right) = 0.5 \mu g \left( \frac{1}{2} \right) = 0.25 \mu g = \frac{1}{4} \mu g$
$R = \frac{1}{4} \frac{m_1 m_2}{m_1 + m_2} \mu g$.

(A) The acceleration $a$ of a block sliding down an inclined plane is given by the formula:
$a = g \sin \theta - \mu g \cos \theta$
Given: mass $m = 10 \ kg$,angle $\theta = 45^{\circ}$,coefficient of kinetic friction $\mu = 0.3$,time $t = 2 \ s$,and acceleration due to gravity $g = 10 \ m/s^2$.
Substituting these values into the acceleration formula:
$a = 10 \sin 45^{\circ} - 0.3 \times 10 \cos 45^{\circ}$
$a = 10 \times \frac{1}{\sqrt{2}} - 3 \times \frac{1}{\sqrt{2}} = \frac{7}{\sqrt{2}} \ m/s^2$
Using the second equation of motion for distance $s$ with initial velocity $u = 0$:
$s = ut + \frac{1}{2}at^2$
$s = 0 \times 2 + \frac{1}{2} \times \left( \frac{7}{\sqrt{2}} \right) \times (2)^2$
$s = \frac{1}{2} \times \frac{7}{\sqrt{2}} \times 4 = \frac{14}{\sqrt{2}} = 7 \sqrt{2} \ m$
Thus,the distance travelled is $7 \sqrt{2} \ m$.

(A) Given that,$\mu = C s^2$.
The net force on the block along the incline is given by:
$M g \sin \theta - f = M a$
$M g \sin \theta - \mu M g \cos \theta = M a$
$a = g \sin \theta - \mu g \cos \theta$
Substituting $\mu = C s^2$ and $\theta = 45^{\circ}$:
$a = g \sin 45^{\circ} - C s^2 g \cos 45^{\circ} = \frac{g}{\sqrt{2}} (1 - C s^2)$
Using the relation $a = v \frac{dv}{ds}$,we have:
$v \frac{dv}{ds} = \frac{g}{\sqrt{2}} (1 - C s^2)$
$v dv = \frac{g}{\sqrt{2}} (1 - C s^2) ds$
Integrating both sides from initial state $(s=0, v=0)$ to final state $(s=s_{max}, v=0)$:
$\int_{0}^{0} v dv = \int_{0}^{s_{max}} \frac{g}{\sqrt{2}} (1 - C s^2) ds$
$0 = \frac{g}{\sqrt{2}} [s - \frac{C s^3}{3}]_{0}^{s_{max}}$
Since $g \neq 0$,we have:
$s_{max} - \frac{C s_{max}^3}{3} = 0$
$s_{max} (1 - \frac{C s_{max}^2}{3}) = 0$
Since $s_{max} \neq 0$,we get:
$1 = \frac{C s_{max}^2}{3}$
$s_{max}^2 = \frac{3}{C}$
$s_{max} = \sqrt{\frac{3}{C}}$

(D) To keep the box in static equilibrium,the downward force along the incline $(mg \sin \theta)$ must be balanced by the maximum static friction force $(f_{max})$.
The normal force $N$ acting on the box is the sum of the component of weight perpendicular to the incline and the applied force $F$:
$N = mg \cos \theta + F$
The maximum static friction force is given by:
$f_{max} = \mu N = \mu(mg \cos \theta + F)$
For equilibrium,$mg \sin \theta \leq f_{max}$,so the minimum force $F$ required is when $mg \sin \theta = \mu(mg \cos \theta + F)$.
Rearranging for $F$:
$F = \frac{mg \sin \theta}{\mu} - mg \cos \theta = mg \left( \frac{\sin \theta}{\mu} - \cos \theta \right)$
Given $m = 1 \ kg$,$g = 10 \ m/s^2$,$\theta = 30^{\circ}$,and $\mu = 0.2$:
$F = 1 \times 10 \left( \frac{\sin 30^{\circ}}{0.2} - \cos 30^{\circ} \right)$
$F = 10 \left( \frac{0.5}{0.2} - \frac{\sqrt{3}}{2} \right) = 10 \left( 2.5 - 0.866 \right)$
$F = 10 \times 1.634 = 16.34 \ N$
Thus,the minimum magnitude of $F$ is $\geq 16.3 \ N$.

(C) For upward motion,the force required is $F_{\text{up}} = mg(\sin \theta + \mu \cos \theta)$.
For downward motion,the force required to prevent sliding is $F_{\text{down}} = mg(\sin \theta - \mu \cos \theta)$.
According to the problem,$F_{\text{up}} = 2 F_{\text{down}}$.
Substituting the expressions,we get $mg(\sin \theta + \mu \cos \theta) = 2mg(\sin \theta - \mu \cos \theta)$.
Dividing both sides by $mg$,we have $\sin \theta + \mu \cos \theta = 2 \sin \theta - 2 \mu \cos \theta$.
Rearranging the terms,$3 \mu \cos \theta = \sin \theta$,which simplifies to $\mu = \frac{1}{3} \tan \theta$.
Given $\theta = 60^{\circ}$,we have $\mu = \frac{1}{3} \tan 60^{\circ} = \frac{1}{3} \times \sqrt{3} = \frac{1}{\sqrt{3}}$.

(A) For a smooth inclined plane,the acceleration is $a_s = g \sin \theta$. The time taken to cover distance $s$ is $t_s = \sqrt{\frac{2s}{g \sin \theta}}$.
For a rough inclined plane,the acceleration is $a_r = g(\sin \theta - \mu \cos \theta)$. The time taken is $t_r = \sqrt{\frac{2s}{g(\sin \theta - \mu \cos \theta)}} = n t_s$.
Squaring both sides: $\frac{2s}{g(\sin \theta - \mu \cos \theta)} = n^2 \frac{2s}{g \sin \theta}$.
This simplifies to $\sin \theta = n^2(\sin \theta - \mu \cos \theta)$,which gives $\mu \cos \theta = \sin \theta (1 - \frac{1}{n^2})$.
Thus,$\mu = \tan \theta (1 - \frac{1}{n^2})$.
Given $\theta = 45^{\circ}$,$\tan 45^{\circ} = 1$,so $\mu = 1 - \frac{1}{n^2}$.

(C) The minimum force required to move a body up a rough inclined plane is given by $F_1 = mg(\sin \theta + \mu \cos \theta)$.
The minimum force required to prevent the body from sliding down the rough inclined plane is given by $F_2 = mg(\sin \theta - \mu \cos \theta)$.
According to the problem,$F_1 = 3F_2$.
Substituting the expressions,we get $mg(\sin \theta + \mu \cos \theta) = 3mg(\sin \theta - \mu \cos \theta)$.
Dividing both sides by $mg$,we have $\sin \theta + \mu \cos \theta = 3\sin \theta - 3\mu \cos \theta$.
Rearranging the terms,$4\mu \cos \theta = 2\sin \theta$,which simplifies to $\tan \theta = 2\mu$.
Given $\mu = \frac{1}{2\sqrt{3}}$,we have $\tan \theta = 2 \times \frac{1}{2\sqrt{3}} = \frac{1}{\sqrt{3}}$.
Therefore,$\theta = \tan^{-1}(\frac{1}{\sqrt{3}}) = 30^{\circ}$.

(C) Given: Coefficient of friction $\mu = 0.5$. The ratio of net force $F$ to normal reaction $R$ is $\frac{F}{R} = \frac{1}{2}$.
The net force $F$ acting on the body sliding down the plane is given by $F = mg \sin \theta - f$,where $f$ is the frictional force.
The frictional force is $f = \mu R$,and the normal reaction is $R = mg \cos \theta$.
Substituting these into the expression for $F$: $F = mg \sin \theta - \mu mg \cos \theta$.
Now,using the given ratio $\frac{F}{R} = \frac{1}{2}$:
$\frac{mg \sin \theta - \mu mg \cos \theta}{mg \cos \theta} = \frac{1}{2}$
Dividing the numerator by the denominator:
$\tan \theta - \mu = \frac{1}{2}$
$\tan \theta = \frac{1}{2} + \mu = 0.5 + 0.5 = 1.0$.
Since $\tan \theta = 1$,we have $\theta = 45^{\circ}$.

(B) The normal force on the block is $N = mg \cos \theta$.
Since the coefficient of friction is $\mu = \frac{2}{3} x$,the kinetic friction force is $f_k = \mu N = \left( \frac{2}{3} x \right) mg \cos \theta$.
According to the work-energy theorem,the total work done on the block is equal to the change in its kinetic energy: $W_g + W_f = \Delta K.E. = 0 - 0 = 0$.
The work done by gravity as the block slides down a distance $x$ is $W_g = mgx \sin \theta$.
The work done by the variable friction force is $W_f = - \int_0^x f_k \, dx = - \int_0^x \left( \frac{2}{3} x mg \cos \theta \right) dx = - \frac{2}{3} mg \cos \theta \left[ \frac{x^2}{2} \right]_0^x = - \frac{1}{3} mgx^2 \cos \theta$.
Substituting these into the work-energy equation: $mgx \sin \theta - \frac{1}{3} mgx^2 \cos \theta = 0$.
Dividing by $mgx$ (assuming $x \neq 0$): $\sin \theta - \frac{1}{3} x \cos \theta = 0$.
Solving for $x$: $x = 3 \tan \theta$.
Given $\theta = 30^{\circ}$,$x = 3 \tan 30^{\circ} = 3 \times \frac{1}{\sqrt{3}} = \sqrt{3} \text{ m}$.

(D) The total retarding force acting on the body moving up the inclined plane is $F_{net} = mg \sin \theta + f_k$,where $f_k = \mu R = \mu mg \cos \theta$.
Thus,$F_{net} = mg(\sin \theta + \mu \cos \theta)$.
The retardation $a$ is given by $a = \frac{F_{net}}{m} = g(\sin \theta + \mu \cos \theta)$.
Using the work-energy theorem,the total work done by all forces equals the change in kinetic energy: $W_{total} = \Delta KE = 0 - E = -E$.
The work done by gravity is $W_g = -mg \sin \theta \cdot s$ and the work done against friction is $W_f = f_k \cdot s = \mu mg \cos \theta \cdot s$.
From $v^2 - u^2 = 2as$,we have $0 - u^2 = -2as$,so $s = \frac{u^2}{2a} = \frac{2E/m}{2g(\sin \theta + \mu \cos \theta)} = \frac{E}{mg(\sin \theta + \mu \cos \theta)}$.
The work done against friction is $W_f = (\mu mg \cos \theta) \cdot s = (\mu mg \cos \theta) \cdot \frac{E}{mg(\sin \theta + \mu \cos \theta)} = \frac{\mu E \cos \theta}{\sin \theta + \mu \cos \theta}$.

(A) For a smooth inclined plane,the acceleration is $a_s = g \sin \theta$. The time taken to cover distance $d$ is $t_s = \sqrt{2d / a_s} = \sqrt{2d / (g \sin \theta)}$.
For a rough inclined plane,the acceleration is $a_r = g(\sin \theta - \mu \cos \theta)$. The time taken to cover distance $d$ is $t_r = \sqrt{2d / a_r} = \sqrt{2d / (g(\sin \theta - \mu \cos \theta))}$.
Given $t_r = t$ and $t_s = t/2$,we have $t_r = 2t_s$.
Therefore,$\sqrt{2d / (g(\sin \theta - \mu \cos \theta))} = 2 \sqrt{2d / (g \sin \theta)}$.
Squaring both sides: $1 / (\sin \theta - \mu \cos \theta) = 4 / \sin \theta$.
$\sin \theta = 4 \sin \theta - 4 \mu \cos \theta$.
$4 \mu \cos \theta = 3 \sin \theta$.
$\mu = (3/4) \tan \theta$.
Substituting $\theta = 45^{\circ}$,$\mu = (3/4) \tan 45^{\circ} = 3/4 \times 1 = 3/4$.

(B) For the block to move down without acceleration,the net force along the inclined plane must be zero.
Let $F$ be the applied force directed up the incline.
The forces acting down the incline are the component of gravity $mg \sin 30^{\circ}$.
The forces acting up the incline are the applied force $F$ and the kinetic friction $f_k = \mu N = \mu mg \cos 30^{\circ}$.
Equating the forces: $mg \sin 30^{\circ} = F + \mu mg \cos 30^{\circ}$.
Substituting the values: $5 \times 10 \times \frac{1}{2} = F + \frac{\sqrt{3}}{2} \times 5 \times 10 \times \frac{\sqrt{3}}{2}$.
$25 = F + \frac{\sqrt{3}}{2} \times 50 \times \frac{\sqrt{3}}{2} = F + \frac{3}{4} \times 50 = F + 37.5$.
$F = 25 - 37.5 = -12.5 \ N$.
The negative sign indicates that the force $F$ must be applied in the opposite direction (i.e.,down the incline) to maintain constant velocity.
Therefore,a force of $12.5 \ N$ must be applied down the incline.

(C) Let $L$ be the length of the incline.
For a smooth plane,the acceleration is $a_1 = g \sin\theta$. The time taken is $t_1 = \sqrt{2L/a_1}$.
For a rough plane,the acceleration is $a_2 = g(\sin\theta - \mu \cos\theta)$. The time taken is $t_2 = \sqrt{2L/a_2}$.
Given that $t_2 = 1.5 t_1$,we have $t_2^2 = 2.25 t_1^2$.
Substituting the expressions for $t_1$ and $t_2$,we get $\frac{2L}{a_2} = 2.25 \frac{2L}{a_1}$,which simplifies to $a_1 = 2.25 a_2$.
Substituting $a_1$ and $a_2$,we get $g \sin\theta = 2.25 g(\sin\theta - \mu \cos\theta)$.
Since $\theta = 45^\circ$,$\sin\theta = \cos\theta = 1/\sqrt{2}$.
Dividing by $g \sin\theta$,we get $1 = 2.25(1 - \mu)$.
Thus,$1 - \mu = 1/2.25 = 4/9$.
Therefore,$\mu = 1 - 4/9 = 5/9$.

(A) Let the length of the inclined plane be $L$. The acceleration of the block on a rough surface is $a_1 = g(\sin \theta - \mu \cos \theta)$ and on a smooth surface is $a_2 = g \sin \theta$.
Using the equation of motion $s = ut + \frac{1}{2}at^2$,with initial velocity $u = 0$,we have $L = \frac{1}{2} a_1 t^2$ and $L = \frac{1}{2} a_2 (t/2)^2$.
Equating the two expressions for $L$: $\frac{1}{2} a_1 t^2 = \frac{1}{2} a_2 \frac{t^2}{4}$,which simplifies to $a_1 = \frac{a_2}{4}$ or $4a_1 = a_2$.
Substituting the expressions for $a_1$ and $a_2$ with $\theta = 45^\circ$ (where $\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}$):
$4g(\frac{1}{\sqrt{2}} - \mu \frac{1}{\sqrt{2}}) = g(\frac{1}{\sqrt{2}})$.
Dividing both sides by $\frac{g}{\sqrt{2}}$,we get $4(1 - \mu) = 1$.
$4 - 4\mu = 1 \implies 4\mu = 3 \implies \mu = 0.75$.
Since $\mu = \frac{\alpha}{100}$,we have $\frac{\alpha}{100} = 0.75$,which gives $\alpha = 75$.