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

(D) Since the block is at rest,the net force acting on it is zero.
The forces acting on the block are its weight $(mg)$,the normal force $(N)$ exerted by the surface,and the static friction force $(f)$ exerted by the surface.
The net force exerted by the surface on the block is the vector sum of the normal force and the friction force,which is given by $F_{surface} = \sqrt{N^2 + f^2}$.
For a block at rest on an inclined plane:
$N = mg \cos \theta$
$f = mg \sin \theta$
Therefore,the net force exerted by the surface is:
$F_{surface} = \sqrt{(mg \cos \theta)^2 + (mg \sin \theta)^2}$
$F_{surface} = \sqrt{m^2g^2(\cos^2 \theta + \sin^2 \theta)}$
$F_{surface} = mg$
Given $m = 8\, kg$ and $g = 10\, m/s^2$:
$F_{surface} = 8 \times 10 = 80\, N$.
Thus,the magnitude of the net force exerted by the surface on the block is $80\, N$.

(B) The acceleration $a$ of a block sliding down an inclined plane with friction is given by $a = g(\sin \theta - \mu \cos \theta)$.
Given: $g = 10\, ms^{-2}$,$\theta = 30^o$,$\mu = 0.2$,and $t = 5\, s$.
Substituting the values:
$a = 10 \times (\sin 30^o - 0.2 \times \cos 30^o)$
$a = 10 \times (0.5 - 0.2 \times 0.866)$
$a = 10 \times (0.5 - 0.1732) = 10 \times 0.3268 = 3.268\, ms^{-2}$.
Using the equation of motion $v = u + at$,where initial velocity $u = 0$:
$v = 0 + 3.268 \times 5 = 16.34\, ms^{-1}$.

(C) The component of the weight of the block acting parallel to the inclined plane is given by $F = mg \sin \theta$,where $m$ is the mass of the block,$g$ is the acceleration due to gravity,and $\theta$ is the angle of inclination.
As the inclined plane is made horizontal,the angle $\theta$ decreases.
Since the sine function is an increasing function for $0^\circ \le \theta \le 90^\circ$,as $\theta$ decreases,$\sin \theta$ also decreases.
Therefore,the component of weight parallel to the plane,$mg \sin \theta$,decreases.

(D) Let $m$ be the mass of the block. The force required to lift the block vertically is $F_1 = mg$.
The force required to drag the block up the inclined plane is $F_2 = mg \sin \theta + f_k$,where $f_k = \mu N = \mu mg \cos \theta$.
So,$F_2 = mg \sin \theta + \mu mg \cos \theta$.
According to the problem,$F_2 < F_1$,therefore:
$mg \sin \theta + \mu mg \cos \theta < mg$.
Dividing by $mg$,we get:
$\sin \theta + \mu \cos \theta < 1$.
Substituting $\theta = 30^o$:
$\sin 30^o + \mu \cos 30^o < 1$.
$1/2 + \mu (\sqrt{3}/2) < 1$.
$\mu (\sqrt{3}/2) < 1 - 1/2$.
$\mu (\sqrt{3}/2) < 1/2$.
$\mu < 1/\sqrt{3}$.

(C) The component of acceleration due to gravity along the slope is $a_g = g \sin \theta = 10 \sin 30^{\circ} = 5\, m/s^2$.
The acceleration due to friction opposing the motion is $a_f = \mu g \cos \theta = 0.5 \times 10 \times \cos 30^{\circ} = 5 \times 0.866 = 4.33\, m/s^2$.
The net acceleration of the car down the slope is $a = a_g - a_f = 5 - 4.33 = 0.67\, m/s^2$.
Using the equation of motion $v^2 = u^2 + 2as$,where $u = 6\, m/s$,$a = 0.67\, m/s^2$,and $s = 15\, m$:
$v^2 = (6)^2 + 2 \times 0.67 \times 15$
$v^2 = 36 + 20.1 = 56.1$
$v = \sqrt{56.1} \approx 7.49\, m/s$.

(A) Let the length of the inclined plane be $s$.
For the smooth inclined plane,the acceleration is $a_1 = g \sin \theta$. Using $v^2 - u^2 = 2as$ with $u=0$,we get $v^2 = 2(g \sin \theta)s$ --- $(1)$.
For the rough inclined plane,the acceleration is $a_2 = g(\sin \theta - \mu \cos \theta)$. The final velocity is $v' = v/n$. Using $v'^2 = 2a_2s$,we get $(v/n)^2 = 2g(\sin \theta - \mu \cos \theta)s$ --- $(2)$.
Dividing equation $(1)$ by equation $(2)$,we get: $n^2 = \frac{\sin \theta}{\sin \theta - \mu \cos \theta}$.
Rearranging the terms: $n^2(\sin \theta - \mu \cos \theta) = \sin \theta$.
$n^2 \sin \theta - n^2 \mu \cos \theta = \sin \theta$.
$n^2 \mu \cos \theta = n^2 \sin \theta - \sin \theta = \sin \theta (n^2 - 1)$.
$\mu = \frac{\sin \theta (n^2 - 1)}{n^2 \cos \theta} = \tan \theta \left( \frac{n^2 - 1}{n^2} \right) = \tan \theta \left( 1 - \frac{1}{n^2} \right)$.

(A) For a smooth inclined plane,the acceleration is $a_2 = g \sin \theta$. The time taken is $t_2 = \sqrt{\frac{2s}{g \sin \theta}}$.
For a rough inclined plane,the acceleration is $a_1 = g(\sin \theta - \mu \cos \theta)$. The time taken is $t_1 = \sqrt{\frac{2s}{g(\sin \theta - \mu \cos \theta)}}$.
Given $t_1 = \frac{4}{3} t_2$,so $\frac{t_1}{t_2} = \frac{4}{3}$.
Squaring both sides,$\frac{t_1^2}{t_2^2} = \frac{16}{9} = \frac{a_2}{a_1} = \frac{g \sin \theta}{g(\sin \theta - \mu \cos \theta)} = \frac{\sin \theta}{\sin \theta - \mu \cos \theta}$.
Since $\theta = 45^{\circ}$,$\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$.
$\frac{16}{9} = \frac{1/\sqrt{2}}{1/\sqrt{2} - \mu(1/\sqrt{2})} = \frac{1}{1 - \mu}$.
$16(1 - \mu) = 9 \implies 16 - 16\mu = 9 \implies 16\mu = 7 \implies \mu = \frac{7}{16}$.

(A) The forces acting on the block of mass $m = 15\,kg$ are:
$1$. Weight $mg = 15 \times 10 = 150\,N$ acting vertically downwards.
$2$. Normal force $N_1$ perpendicular to the inclined plane.
$3$. Tension $T = 50\,N$ acting horizontally.
$4$. Friction force $f$ acting up the incline.
Resolving forces perpendicular to the inclined plane:
$N_1 = mg \cos(45^{\circ}) + T \sin(45^{\circ})$
$N_1 = 150 \times \frac{1}{\sqrt{2}} + 50 \times \frac{1}{\sqrt{2}} = \frac{200}{\sqrt{2}}\,N$
Resolving forces parallel to the inclined plane:
$f + T \cos(45^{\circ}) = mg \sin(45^{\circ})$
$f = 150 \times \frac{1}{\sqrt{2}} - 50 \times \frac{1}{\sqrt{2}} = \frac{100}{\sqrt{2}}\,N$
Using the relation $f = \mu N_1$:
$\frac{100}{\sqrt{2}} = \mu \times \frac{200}{\sqrt{2}}$
$\mu = \frac{100}{200} = \frac{1}{2}$

(C) Let the total length of the rope be $L$ and its total mass be $M$. The mass per unit length is $\lambda = M/L$.
The length of the hanging part is $l = 0.25L = L/4$. The mass of the hanging part is $m = \lambda l = M/4$.
The length of the part on the table is $L - l = 0.75L = 3L/4$. The mass of this part is $M' = \lambda(3L/4) = 3M/4$.
The downward force causing the rope to slide is the weight of the hanging part: $F_g = mg = (M/4)g$.
The limiting friction force opposing the motion is $f_s = \mu_s N = \mu_s M' g = \mu_s (3M/4)g$.
At the point of sliding,the downward force equals the limiting friction force: $(M/4)g = \mu_s (3M/4)g$.
Solving for $\mu_s$: $\mu_s = (M/4) / (3M/4) = 1/3 \approx 0.33$.

(A) The forces acting on the insect are the gravitational force $mg$ acting vertically downwards,the normal reaction $N$ acting radially outwards,and the frictional force $f$ acting tangentially upwards along the surface.
Resolving the gravitational force $mg$ into two components:
$1$. $mg \cos \alpha$ acting radially inwards (opposite to $N$).
$2$. $mg \sin \alpha$ acting tangentially downwards along the surface.
Since the insect crawls very slowly,it is in equilibrium. Thus:
$N = mg \cos \alpha$
For the insect to not slip,the frictional force $f$ must balance the tangential component of gravity:
$f = mg \sin \alpha$
We know that the maximum static friction is $f_{max} = \mu N$.
At the maximum angle $\alpha$,$f = f_{max} = \mu N$.
Substituting the expressions for $f$ and $N$:
$mg \sin \alpha = \mu (mg \cos \alpha)$
Dividing both sides by $mg \cos \alpha$:
$\tan \alpha = \mu$
Given $\mu = 1/3$,we have:
$\tan \alpha = 1/3$
Therefore,$\cot \alpha = 1/\tan \alpha = 3$.

(B) Let $\theta$ be the angle the radius makes with the vertical at the maximum height reached by the insect.
At this point,the component of gravity along the tangent balances the limiting friction.
The forces acting on the insect are: gravity $(mg)$ acting downwards,normal reaction $(N)$ perpendicular to the surface,and friction $(f)$ acting upwards along the tangent.
Resolving gravity into components: $mg \sin \theta$ along the tangent and $mg \cos \theta$ perpendicular to the surface.
Thus,$N = mg \cos \theta$.
The limiting friction is $f = \mu N = \mu mg \cos \theta$.
For the insect to be in equilibrium,$f = mg \sin \theta$.
Therefore,$\mu mg \cos \theta = mg \sin \theta$,which gives $\tan \theta = \mu$.
From the geometry of the bowl,the height $h$ from the bottom is $h = r - r \cos \theta = r(1 - \cos \theta)$.
Since $\tan \theta = \mu$,we have $\cos \theta = \frac{1}{\sqrt{1+\mu^2}}$.
Substituting this into the expression for $h$,we get $h = r\left[1 - \frac{1}{\sqrt{1+\mu^2}}\right]$.

(A) The forces acting on the block along the inclined plane are the component of gravitational force $mg \sin 30^{\circ}$ acting downwards,the tension $T$ in the string acting upwards,and the frictional force $f$ acting upwards.
Maximum static frictional force $f_{max} = \mu N = \mu mg \cos 30^{\circ}$.
Given $\mu = 0.8$,$m = 1 \text{ kg}$,$g = 9.8 \text{ m/s}^2$,and $\theta = 30^{\circ}$.
$f_{max} = 0.8 \times 1 \times 9.8 \times \cos 30^{\circ} = 0.8 \times 9.8 \times 0.866 \approx 6.79 \text{ N}$.
The downward force component is $F_{down} = mg \sin 30^{\circ} = 1 \times 9.8 \times 0.5 = 4.9 \text{ N}$.
Since the downward force $F_{down} = 4.9 \text{ N}$ is less than the maximum static frictional force $f_{max} \approx 6.79 \text{ N}$,the block has no tendency to slide down the incline.
Therefore,the static friction alone is sufficient to balance the downward component of gravity,and the tension in the string is $T = 0 \text{ N}$.

(B) Let $\lambda$ be the mass per unit length of the chain.
Let $l$ be the length of the chain hanging over the edge of the table.
The length of the chain remaining on the table is $(L - l)$.
The force pulling the chain down is the weight of the hanging part: $F_g = l \lambda g$.
The maximum static frictional force acting on the part of the chain on the table is: $f_{max} = \mu N = \mu (L - l) \lambda g$.
For the chain to be in limiting equilibrium,the downward force must equal the maximum frictional force:
$l \lambda g = \mu (L - l) \lambda g$
$l = \mu (L - l)$
$l = \mu L - \mu l$
$l(1 + \mu) = \mu L$
$\frac{l}{L} = \frac{\mu}{\mu + 1}$
Thus,the maximum fraction of the chain that can hang is $\frac{\mu}{\mu + 1}$.

(A) The net force acting on the mass along the inclined plane is $F_{net} = mg \sin \theta - f_r = ma$.
Since the frictional force is $f_r = \mu N = \mu mg \cos \theta$,we have $mg \sin \theta - \mu mg \cos \theta = ma$.
Substituting $\mu = \mu_0 x$,the acceleration $a$ is given by $a = g(\sin \theta - \mu_0 x \cos \theta)$.
Using the relation $a = v \frac{dv}{dx}$,we integrate the equation of motion: $\int_0^v v \, dv = \int_0^x g(\sin \theta - \mu_0 x \cos \theta) \, dx$.
At the point where the mass stops,the final velocity $v = 0$.
Thus,$0 = g [x \sin \theta - \frac{\mu_0 x^2}{2} \cos \theta]$.
Setting the term inside the bracket to zero: $x \sin \theta = \frac{\mu_0 x^2}{2} \cos \theta$.
Solving for $x$,we get $x = \frac{2 \sin \theta}{\mu_0 \cos \theta} = \frac{2}{\mu_0} \tan \theta$.

(A) The acceleration of the mass is given by $a = g(\sin \theta - \mu_0 x \cos \theta)$.
Using the work-energy theorem or integrating $v dv = a dx$,we have $\int_0^{v} v dv = \int_0^{x/2} g(\sin \theta - \mu_0 x \cos \theta) dx$.
Integrating both sides: $\frac{v^2}{2} = g[\sin \theta \cdot x' - \frac{\mu_0}{2} (x')^2 \cos \theta]$,where $x' = x/2$.
Substituting $x' = \frac{x}{2} = \frac{1}{\mu_0} \tan \theta$ (since maximum distance $x_{max} = \frac{2 \tan \theta}{\mu_0}$):
$\frac{v^2}{2} = g[\sin \theta (\frac{\tan \theta}{\mu_0}) - \frac{\mu_0}{2} (\frac{\tan \theta}{\mu_0})^2 \cos \theta]$.
$\frac{v^2}{2} = g[\frac{\sin \theta \tan \theta}{\mu_0} - \frac{\tan^2 \theta \cos \theta}{2 \mu_0}] = g[\frac{\sin \theta \tan \theta}{\mu_0} - \frac{\sin \theta \tan \theta}{2 \mu_0}] = \frac{g \sin \theta \tan \theta}{2 \mu_0}$.
Thus,$v^2 = \frac{g \sin \theta \tan \theta}{\mu_0}$,which gives $v = \sqrt{\frac{g \tan \theta \sin \theta}{\mu_0}}$.

(D) For a body to be in equilibrium on an inclined plane, the component of gravitational force acting down the plane must be balanced by the static frictional force.
The force acting down the plane is $F_g = mg \sin \theta$.
The maximum static frictional force is $f_{s,max} = \mu_s N$, where $N = mg \cos \theta$ is the normal force.
For equilibrium, the condition is $mg \sin \theta \leq \mu_s mg \cos \theta$.
This simplifies to $\mu_s \geq \tan \theta$.
Given the inclination $\theta = 45^{\circ}$, we have $\mu_s \geq \tan 45^{\circ}$.
Since $\tan 45^{\circ} = 1$, the condition becomes $\mu_s \geq 1$.

(A) The acceleration $a$ of a body sliding down an inclined plane with friction is given by the formula: $a = g \sin \theta - \mu g \cos \theta$.
Given: $g = 9.8\, m/s^2$,$\theta = 45^{\circ}$,and $\mu = 0.5$.
Substituting the values:
$a = 9.8 \sin 45^{\circ} - 0.5 \times 9.8 \cos 45^{\circ}$
Since $\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$:
$a = 9.8 \left( \frac{1}{\sqrt{2}} \right) - 0.5 \times 9.8 \left( \frac{1}{\sqrt{2}} \right)$
$a = 9.8 \left( \frac{1}{\sqrt{2}} \right) (1 - 0.5)$
$a = 9.8 \times 0.5 \times \frac{1}{\sqrt{2}}$
$a = 4.9 \times \frac{1}{\sqrt{2}} = \frac{4.9}{\sqrt{2}}\, m/s^2$.

(B) When block $A$ slides down a rough inclined plane,work is done against the frictional force,which converts some of its potential energy into heat energy.
As a result,the kinetic energy of block $A$ at the bottom is less than its initial potential energy.
In contrast,block $B$ falls freely under gravity,meaning its potential energy is entirely converted into kinetic energy (ignoring air resistance).
Therefore,the final velocity of block $B$ will be greater than that of block $A$ upon reaching the ground.

(A) Case $I$: As the block slides down with constant velocity,the acceleration is zero. In this case,the frictional force $f$ balances the component of gravity along the plane:
$f = mg \sin \theta$ and $f = \mu R = \mu mg \cos \theta$
Therefore,$\mu mg \cos \theta = mg \sin \theta$,which gives $\mu = \tan \theta$ ... $(i)$
Case $II$: The block is projected upward with initial velocity $u$. It experiences a downward acceleration $a$ due to both gravity and friction acting down the plane:
$mg \sin \theta + f = ma$
$mg \sin \theta + \mu mg \cos \theta = ma$
Substituting $\mu = \tan \theta$ from $(i)$:
$mg \sin \theta + (\tan \theta) mg \cos \theta = ma$
$mg \sin \theta + mg \sin \theta = ma$
$2mg \sin \theta = ma \implies a = 2g \sin \theta$ ... $(ii)$
Let $x$ be the distance moved up the plane before the block comes to rest. Using the equation of motion $v^2 - u^2 = 2as$:
$0^2 - u^2 = 2(-a)x$
$-u^2 = -2(2g \sin \theta)x$
$x = \frac{u^2}{4g \sin \theta}$

(B) The block is at rest on the inclined plane. This means the net force acting on the block along the inclined plane must be zero.
The gravitational force acting on the block is $mg$,which can be resolved into two components:
$1$. $A$ component perpendicular to the inclined plane: $mg \cos \theta$.
$2$. $A$ component parallel to the inclined plane: $mg \sin \theta$.
The component $mg \sin \theta$ acts downwards along the plane,trying to slide the block down. Since the block is at rest,the static frictional force $f$ must balance this component to prevent motion.
Therefore,the frictional force $f = mg \sin \theta$.

(A) To move the block up the inclined plane,the applied force $F_1$ must overcome both the component of gravitational force acting down the plane and the maximum static frictional force acting down the plane.
$1$. The component of gravitational force acting down the inclined plane is $mg \sin \theta$.
$2$. The normal force acting on the block is $N = mg \cos \theta$.
$3$. The maximum static frictional force acting down the plane is $f = \mu N = \mu mg \cos \theta$.
$4$. Therefore,the total force required to move the block up the plane is $F_1 = mg \sin \theta + f = mg \sin \theta + \mu mg \cos \theta = mg(\sin \theta + \mu \cos \theta)$.

(C) Let $M$ be the total mass of the chain of length $L$. The mass per unit length is $\lambda = \frac{M}{L}$.
When a length $l$ hangs from the table,the mass of the hanging part is $m_1 = \lambda l = \frac{M}{L} l$.
The mass of the part remaining on the table is $m_2 = \lambda (L - l) = \frac{M}{L} (L - l)$.
For the chain to be in equilibrium,the downward force due to the hanging part must be balanced by the maximum static frictional force acting on the part on the table.
The downward force is $F_g = m_1 g = \frac{M}{L} l g$.
The maximum frictional force is $f_{max} = \mu N = \mu m_2 g = \mu \frac{M}{L} (L - l) g$.
Equating the two: $\frac{M}{L} l g = \mu \frac{M}{L} (L - l) g$.
Solving for $\mu$: $\mu = \frac{l}{L - l}$.

(B) According to the Work-Energy Theorem,the total work done on the bar is equal to the change in its kinetic energy.
Since the bar starts from rest and stops at distance $S$,the change in kinetic energy $\Delta K = 0 - 0 = 0$.
Therefore,the total work done $W = \int_{0}^{S} (mg \sin \theta - \mu mg \cos \theta) dx = 0$.
Substituting $\mu = kx$,we get $\int_{0}^{S} (mg \sin \theta - kx mg \cos \theta) dx = 0$.
Integrating with respect to $x$: $[mg \sin \theta \cdot x - \frac{1}{2} k mg \cos \theta \cdot x^2]_0^S = 0$.
This simplifies to $mg \sin \theta \cdot S = \frac{1}{2} k mg \cos \theta \cdot S^2$.
Dividing both sides by $mg \cos \theta \cdot S$ (assuming $S \neq 0$),we get $\tan \theta = \frac{1}{2} k S$.
Solving for $S$,we find $S = \frac{2 \tan \theta}{k}$.

(D) Let $m$ be the mass of the body and $\theta$ be the angle of inclination.
$1$. Force required to just prevent the body from sliding down $(F_1)$:
The component of gravity acting down the plane is $mg \sin \theta$. The friction force acts up the plane to oppose sliding,$f = \mu mg \cos \theta$.
Thus,$F_1 = mg \sin \theta - \mu mg \cos \theta$.
$2$. Force required to just move the body up the plane $(F_2)$:
The component of gravity acts down the plane,and friction also acts down the plane to oppose motion,$f = \mu mg \cos \theta$.
Thus,$F_2 = mg \sin \theta + \mu mg \cos \theta$.
$3$. Given condition: $F_2 = 2 F_1$.
$mg \sin \theta + \mu mg \cos \theta = 2(mg \sin \theta - \mu mg \cos \theta)$
$\sin \theta + \mu \cos \theta = 2 \sin \theta - 2 \mu \cos \theta$
$3 \mu \cos \theta = \sin \theta$
$\tan \theta = 3 \mu$
$\theta = \tan^{-1}(3 \mu)$.

(C) The net force acting on the body along the inclined plane is given by $F_{net} = mg \sin \theta - f_k$,where $f_k = \mu N = \mu mg \cos \theta$.
Since $\mu = 0.3x$,the net force is $F_{net} = mg \sin \theta - (0.3x) mg \cos \theta$.
The acceleration of the body is $a = \frac{F_{net}}{m} = g \sin \theta - 0.3x g \cos \theta$.
The speed of the body increases as long as the acceleration is positive. The speed is maximum when the acceleration becomes zero.
Setting $a = 0$:
$g \sin 37^{\circ} - 0.3x g \cos 37^{\circ} = 0$
Given $\sin 37^{\circ} = 0.6$ and $\cos 37^{\circ} = 0.8$:
$10 \times 0.6 - 0.3 \times x \times 10 \times 0.8 = 0$
$6 - 2.4x = 0$
$2.4x = 6$
$x = \frac{6}{2.4} = 2.5 \, m$.

(A) The condition for a block to just begin sliding down an inclined plane is given by the angle of repose,$\theta$,where $\tan \theta = \mu_s$ (coefficient of static friction).
The coefficient of static friction $\mu_s$ depends only on the nature of the surfaces in contact and is independent of the surface area of the block.
Since the mass and the nature of the surfaces remain unchanged,the coefficient of static friction $\mu_s$ remains the same.
Therefore,the angle of inclination at which the block begins to slide remains $30^{\circ}$.

(C) The forces acting on the block along the incline are the component of gravity $mg \sin \theta$ acting downwards and the kinetic friction force $f_k = \mu_k N = \mu_k mg \cos \theta$ acting upwards.
According to Newton's second law,the net force is $F_{net} = ma$.
$mg \sin 30^o - \mu_k mg \cos 30^o = m \left( \frac{g}{4} \right)$.
Dividing both sides by $mg$,we get $\sin 30^o - \mu_k \cos 30^o = \frac{1}{4}$.
Substituting the values $\sin 30^o = \frac{1}{2}$ and $\cos 30^o = \frac{\sqrt{3}}{2}$,we have $\frac{1}{2} - \mu_k \frac{\sqrt{3}}{2} = \frac{1}{4}$.
Rearranging the terms: $\mu_k \frac{\sqrt{3}}{2} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$.
Solving for $\mu_k$: $\mu_k = \frac{1}{4} \times \frac{2}{\sqrt{3}} = \frac{1}{2\sqrt{3}}$.

(B) The angle of repose $\theta$ is defined as the angle of the inclined plane at which an object placed on it just begins to slide.
At this angle,the component of gravitational force down the plane equals the maximum static frictional force.
Mathematically,$mg \sin \theta = \mu_s mg \cos \theta$.
This simplifies to $\tan \theta = \mu_s$.
Given $\mu_s = \frac{3}{4}$,we have $\tan \theta = 0.75$.
Therefore,$\theta = \tan^{-1}(0.75) = 37^o$.

(A) The contact force $F_c$ between the block and the plane is the resultant of the normal force $N$ and the frictional force $f$.
$F_c = \sqrt{N^2 + f^2}$.
For a block on an inclined plane,the normal force is $N = mg \cos \theta$.
If the block is in equilibrium,the frictional force $f$ balances the component of gravity along the plane,so $f = mg \sin \theta$.
Thus,$F_c = \sqrt{(mg \cos \theta)^2 + (mg \sin \theta)^2} = \sqrt{m^2g^2(\cos^2 \theta + \sin^2 \theta)} = mg$.
Since the contact force $F_c = mg$ is independent of the angle $\theta$,it remains constant as $\theta$ increases from $0^o$ to $90^o$.

(B) Let the total mass of the board and the added weight be $M_{total} = (M + m)$.
When the board is tilted at an angle $\theta$ with the horizontal,the force of gravity $(M + m)g$ acting on the center of mass can be resolved into two components:
$1$. $A$ component perpendicular to the board: $(M + m)g \cos \theta$,which is balanced by the normal reaction $N$.
$2$. $A$ component parallel to the board: $(M + m)g \sin \theta$,which tends to make the board slide down.
For the board to be in equilibrium,the frictional force $f$ must balance the component of gravity acting parallel to the board.
Thus,$f = (M + m)g \sin \theta$.
Since the board is on the verge of sliding,the frictional force is the limiting friction,$f = \mu N$.
Substituting $N = (M + m)g \cos \theta$,we get:
$f = \mu (M + m)g \cos \theta$.
Equating the two expressions for $f$:
$(M + m)g \sin \theta = \mu (M + m)g \cos \theta$
$\mu = \frac{\sin \theta}{\cos \theta} = \tan \theta$.
Therefore,the coefficient of friction is $\tan \theta$.

(A) The acceleration $a$ of a block sliding down a rough inclined plane is given by:
$a = g \sin \theta - \mu g \cos \theta = g(\sin \theta - \mu \cos \theta)$
Given: $u = 0$,$s = 8 \ m$,$t = 2 \ s$,$\theta = 30^{\circ}$,$g = 10 \ m/s^2$.
Using the equation of motion $s = ut + \frac{1}{2}at^2$:
$8 = 0 \times 2 + \frac{1}{2} \times a \times (2)^2$
$8 = 2a \implies a = 4 \ m/s^2$
Now,substitute the expression for $a$:
$4 = 10(\sin 30^{\circ} - \mu \cos 30^{\circ})$
$4 = 10(\frac{1}{2} - \mu \frac{\sqrt{3}}{2})$
$4 = 5 - 5\sqrt{3}\mu$
$5\sqrt{3}\mu = 5 - 4 = 1$
$\mu = \frac{1}{5\sqrt{3}}$

(B) For a smooth incline,the acceleration is $a_1 = g \sin \theta$. The time taken to cover distance '$d$' is $t_1 = \sqrt{\frac{2d}{g \sin \theta}}$.
For a rough incline,the acceleration is $a_2 = g \sin \theta - \mu_k g \cos \theta$. The time taken to cover distance '$d$' is $t_2 = \sqrt{\frac{2d}{g \sin \theta - \mu_k g \cos \theta}}$.
Given $t_2 = n t_1$,we have $n = \frac{t_2}{t_1} = \sqrt{\frac{g \sin \theta}{g \sin \theta - \mu_k g \cos \theta}} = \sqrt{\frac{\sin \theta}{\sin \theta - \mu_k \cos \theta}}$.
Squaring both sides: $n^2 = \frac{\sin \theta}{\sin \theta - \mu_k \cos \theta}$.
Since $\theta = 45^\circ$,$\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}$.
$n^2 = \frac{1/\sqrt{2}}{1/\sqrt{2} - \mu_k (1/\sqrt{2})} = \frac{1}{1 - \mu_k}$.
Rearranging gives $1 - \mu_k = \frac{1}{n^2}$,so $\mu_k = 1 - \frac{1}{n^2}$.

(A) The maximum value of static friction up to which a body does not move is called limiting friction.
Angle of repose $(\alpha)$ is defined as the angle of the inclined plane with the horizontal such that a body placed on it just begins to slide.
In the limiting condition, the forces are balanced:
$F = mg \sin \alpha$
$R = mg \cos \alpha$
Dividing these equations, we get:
$\frac{F}{R} = \tan \alpha$
Since the coefficient of static friction $\mu_s = \frac{F}{R} = \tan \theta$, where $\theta$ is the angle of friction, we have:
$\tan \theta = \tan \alpha \implies \theta = \alpha$
Thus, the angle of repose is equal to the angle of friction. The reason correctly explains the concept of limiting friction used to derive this equality.

(A) Both $Assertion$ and $Reason$ are true and $Reason$ is the correct explanation of $Assertion$. If mountain roads were to go straight up,the angle of inclination $\theta$ would be very large. The frictional force available to prevent slipping is given by $f = \mu mg \cos \theta$. As $\theta$ increases,$\cos \theta$ decreases,making the frictional force smaller. Due to this reduced friction,the vehicle's wheels would be more likely to slip. Furthermore,climbing a steep slope requires significantly higher engine power,which is impractical for most vehicles. Thus,roads are built with a winding path to reduce the effective slope.

(C) Using the third equation of motion,$v^{2} = u^{2} - 2as$. At the highest point,the final velocity $v = 0$.
Therefore,$0 = u^{2} - 2as$,which gives $s = \frac{u^{2}}{2a}$.
For an object on a smooth inclined plane,the acceleration $a = g \sin \theta$.
Thus,the distance traveled is $s = \frac{u^{2}}{2g \sin \theta}$.
Since $u$ and $g$ are constant,$s \propto \frac{1}{\sin \theta}$.
Therefore,$\frac{x_{1}}{x_{2}} = \frac{\sin \theta_{2}}{\sin \theta_{1}} = \frac{\sin 30^{\circ}}{\sin 60^{\circ}}$.
Substituting the values,$\frac{x_{1}}{x_{2}} = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}}$.
Hence,the ratio $x_{1}: x_{2}$ is $1: \sqrt{3}$.

(B) For the box to move,the applied force $F$ must be equal to the limiting friction force:
$F = \mu mg \dots (1)$
For the box not to topple,the torque about the front edge must be zero or balanced. The force $F$ is applied at a height $h = \frac{a}{2} + b$ from the base. The normal force $N$ shifts to the front edge to prevent toppling. Taking torque about the front edge:
$F \left( \frac{a}{2} + b \right) = mg \left( \frac{a}{2} \right) \dots (2)$
Substituting $F = \mu mg$ from equation $(1)$ into equation $(2)$:
$\mu mg \left( \frac{a}{2} + b \right) = mg \left( \frac{a}{2} \right)$
$\mu \left( \frac{a}{2} + b \right) = \frac{a}{2}$
Given $\mu = 0.4 = \frac{2}{5}$,we have:
$\frac{2}{5} \left( \frac{a}{2} + b \right) = \frac{a}{2}$
Multiply by $5$:
$2 \left( \frac{a}{2} + b \right) = 2.5a$
$a + 2b = 2.5a$
$2b = 1.5a$
$\frac{b}{a} = \frac{1.5}{2} = 0.75$
Therefore,$100 \times \frac{b}{a} = 100 \times 0.75 = 75$.

(D) The forces acting on a block of mass $m$ at rest on an inclined plane are:
$(i)$ The weight $mg$ acting vertically downwards.
$(ii)$ The normal force $N$ of the plane on the block.
$(iii)$ The static frictional force $f_{s}$ opposing the impending motion.
In equilibrium,the resultant of these forces must be zero. Resolving the weight $mg$ along the two directions,we have:
$mg \sin \theta = f_{s}$
$mg \cos \theta = N$
As $\theta$ increases,the self-adjusting frictional force $f_{s}$ increases until at $\theta = \theta_{\max}$,$f_{s}$ achieves its maximum value,$(f_{s})_{\max} = \mu_{s} N$.
Therefore,$\tan \theta_{\max} = \mu_{s}$.
When $\theta$ becomes just a little more than $\theta_{\max}$,the block begins to slide.
For $\theta_{\max} = 15^{\circ}$,the coefficient of static friction is:
$\mu_{s} = \tan 15^{\circ} \approx 0.2679 \approx 0.27$.

(N/A) For a vehicle to be parked on a banked slope without slipping,the gravitational force component acting down the slope must be balanced by the static frictional force acting up the slope.
Let $m$ be the mass of the vehicle,$\theta$ be the angle of banking,$\mu_s$ be the coefficient of static friction,and $g$ be the acceleration due to gravity.
The component of weight acting down the slope is $mg \sin \theta$.
The maximum static frictional force available is $f_{s, \max} = \mu_s N$,where $N$ is the normal force.
On a slope,the normal force is $N = mg \cos \theta$.
For the vehicle to remain stationary (parked),the condition is $mg \sin \theta \leq \mu_s mg \cos \theta$.
Dividing both sides by $mg \cos \theta$,we get $\tan \theta \leq \mu_s$.
Thus,the vehicle can be parked on a slope without slipping if $\mu_s \geq \tan \theta$.

(D) For a smooth inclined plane:
The acceleration is $a = g \sin \theta$.
Given $\theta = 45^{\circ}$,so $a = g \sin 45^{\circ} = \frac{g}{\sqrt{2}}$.
The distance $d$ covered in time $T$ starting from rest is $d = \frac{1}{2} a T^2 = \frac{1}{2} \left( \frac{g}{\sqrt{2}} \right) T^2 = \frac{g T^2}{2\sqrt{2}}$.
For a rough inclined plane:
The net force is $F_{net} = mg \sin \theta - f_k = mg \sin \theta - \mu mg \cos \theta = mg(\sin \theta - \mu \cos \theta)$.
The acceleration is $a' = g(\sin \theta - \mu \cos \theta)$.
Since $\theta = 45^{\circ}$,$\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$,so $a' = g \left( \frac{1 - \mu}{\sqrt{2}} \right)$.
The distance $d$ covered in time $pT$ is $d = \frac{1}{2} a' (pT)^2 = \frac{1}{2} g \left( \frac{1 - \mu}{\sqrt{2}} \right) p^2 T^2$.
Equating the two expressions for $d$:
$\frac{g T^2}{2\sqrt{2}} = \frac{g p^2 T^2 (1 - \mu)}{2\sqrt{2}}$
$1 = p^2 (1 - \mu)$
$1 - \mu = \frac{1}{p^2}$
$\mu = 1 - \frac{1}{p^2} = \frac{p^2 - 1}{p^2}$.

(N/A) Consider the diagram shown in the figure.
$(a)$ For the box to just start sliding down the slope:
$f = mg \sin \theta$
$N = mg \cos \theta$
Since $\mu = \frac{f}{N}$,we have $\mu = \frac{mg \sin \theta}{mg \cos \theta} = \tan \theta$.
Therefore,$\theta = \tan^{-1}(\mu)$.
$(b)$ When the angle of inclination is increased to $\alpha > \theta$,the net force $F_1$ acting down the plane is the component of gravity minus the kinetic friction:
$F_1 = mg \sin \alpha - f_k = mg \sin \alpha - \mu N = mg \sin \alpha - \mu mg \cos \alpha = mg(\sin \alpha - \mu \cos \alpha)$.
$(c)$ To keep the box stationary or moving with uniform speed upward,the applied force $F_2$ must overcome both the component of gravity and the friction force (which now acts downwards):
$F_2 = mg \sin \alpha + f_k = mg \sin \alpha + \mu mg \cos \alpha = mg(\sin \alpha + \mu \cos \alpha)$.
$(d)$ When the box is to be moved with acceleration $a$ upward along the plane,the net force $F_3$ must account for the acceleration:
$F_3 = mg(\sin \alpha + \mu \cos \alpha) + ma$.

(N/A) Yes,there is a static friction force $f$ acting on the block.
Since the block is at rest relative to the inclined plane,the force of friction is given by $f = Mg \sin \theta$.
Work done by a force is defined as $W = \vec{F} \cdot \vec{d}$. Since the block is not in motion relative to the inclined plane (displacement $d = 0$),the work done by the friction force is zero.
Because the work done is zero and there is no relative motion between the surfaces,there is no dissipation of energy.

(C) According to the work-energy theorem,the total work done by all forces on the block is equal to the change in its kinetic energy.
Since the block starts from rest at $B$ and comes to rest at $A$,the change in kinetic energy is $\Delta K = 0 - 0 = 0$.
The forces acting on the block are gravity and friction.
Work done by gravity $(W_g)$ = $mg \sin \theta \times (BC + AC)$.
Work done by friction $(W_f)$ = $-\mu mg \cos \theta \times AC$ (since friction only acts on the rough section $CA$).
Setting the total work to zero: $W_g + W_f = 0$.
$mg \sin \theta (BC + AC) - \mu mg \cos \theta (AC) = 0$.
Given $BC = 2AC$,substitute this into the equation:
$mg \sin \theta (2AC + AC) = \mu mg \cos \theta (AC)$.
$3mg \sin \theta (AC) = \mu mg \cos \theta (AC)$.
$3 \sin \theta = \mu \cos \theta$.
$\mu = 3 \tan \theta$.
Comparing this with $\mu = k \tan \theta$,we get $k = 3$.

(D) Let the distance moved up the incline be $s$. The acceleration while moving up is $a_{up} = g(\sin \theta + \mu \cos \theta)$.
Using $v^2 - u^2 = 2as$,for the upward journey: $0 - v_{0}^2 = -2 a_{up} s \implies s = \frac{v_{0}^2}{2g(\sin \theta + \mu \cos \theta)}$.
The acceleration while moving down is $a_{down} = g(\sin \theta - \mu \cos \theta)$.
For the downward journey: $(\frac{v_{0}}{2})^2 - 0 = 2 a_{down} s \implies s = \frac{v_{0}^2}{8g(\sin \theta - \mu \cos \theta)}$.
Equating the two expressions for $s$: $\frac{v_{0}^2}{2g(\sin \theta + \mu \cos \theta)} = \frac{v_{0}^2}{8g(\sin \theta - \mu \cos \theta)}$.
$4(\sin \theta - \mu \cos \theta) = \sin \theta + \mu \cos \theta \implies 3 \sin \theta = 5 \mu \cos \theta$.
$\mu = \frac{3}{5} \tan 30^{\circ} = \frac{3}{5} \times \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{5} \approx 0.3464$.
Given $\mu = \frac{I}{1000}$,we have $I = 346.4$. The nearest integer is $346$.

(D) Let $\theta$ be the angle the radius makes with the vertical at the point where the insect starts slipping. At this point,the component of gravity along the tangent balances the maximum static friction.
$mg \sin \theta = f_{max} = \mu N$
Since the normal force $N = mg \cos \theta$,we have:
$mg \sin \theta = \mu mg \cos \theta$
$\tan \theta = \mu = 0.75 = \frac{3}{4}$
From the geometry of the hemisphere,the height $h$ from the bottom is given by:
$h = R - R \cos \theta = R(1 - \cos \theta)$
Given $\tan \theta = \frac{3}{4}$,we have $\sin \theta = \frac{3}{5}$ and $\cos \theta = \frac{4}{5}$.
Substituting the values:
$h = 1 \times (1 - \frac{4}{5}) = 1 \times \frac{1}{5} = 0.2\, m$.

(C) Given,$\theta=45^{\circ}, s_{1}=s_{2}, u=0$.
On the rough incline,the acceleration is $a_{1}=g(\sin \theta-\mu \cos \theta)$.
On the frictionless incline,the acceleration is $a_{2}=g \sin \theta$.
Let $t_{1}$ be the time taken on the rough plane and $t_{2}$ be the time taken on the frictionless plane. Given $t_{1}=2 t_{2}$.
Using the equation of motion $s=ut+\frac{1}{2}at^{2}$,for $u=0$:
$s_{1}=\frac{1}{2}g(\sin \theta-\mu \cos \theta)t_{1}^{2}$
$s_{2}=\frac{1}{2}g \sin \theta t_{2}^{2}$
Since $s_{1}=s_{2}$,we have:
$\frac{1}{2}g(\sin \theta-\mu \cos \theta)t_{1}^{2}=\frac{1}{2}g \sin \theta t_{2}^{2}$
$\frac{\sin \theta-\mu \cos \theta}{\sin \theta}=\frac{t_{2}^{2}}{t_{1}^{2}}$
Substituting $t_{1}=2t_{2}$:
$1-\mu \cot \theta=\frac{t_{2}^{2}}{(2t_{2})^{2}}=\frac{1}{4}$
$\mu \cot \theta=1-\frac{1}{4}=\frac{3}{4}$
$\mu=\frac{3}{4 \cot \theta}$. Since $\theta=45^{\circ}$,$\cot 45^{\circ}=1$,so $\mu=\frac{3}{4}=0.75$.

(A) When a body slides down an inclined plane with a constant velocity,the net force acting on it is zero.
This implies that the component of gravitational force acting down the plane is balanced by the frictional force acting up the plane.
Let $m$ be the mass,$\theta$ be the angle of inclination,and $\mu$ be the coefficient of friction.
The force down the plane is $mg \sin \theta$.
The frictional force is $f = \mu N = \mu mg \cos \theta$.
Equating the two,we get $mg \sin \theta = \mu mg \cos \theta$.
Therefore,$\mu = \frac{\sin \theta}{\cos \theta} = \tan \theta$.
Given $\theta = 30^{\circ}$,we have $\mu = \tan 30^{\circ} = \frac{1}{\sqrt{3}}$.

(B) The condition for a block to remain stationary on an inclined plane is that the angle of inclination $\theta$ must be less than or equal to the angle of repose $\alpha$,where $\tan \alpha = \mu$.
At the maximum height,the block is at the verge of slipping,so the slope of the tangent to the curve at that point is equal to the coefficient of friction.
$\tan \theta = \frac{dy}{dx} = \mu$
Given $y = \frac{x^2}{4}$,we have $\frac{dy}{dx} = \frac{2x}{4} = \frac{x}{2}$.
Setting $\frac{x}{2} = \mu = 0.5$,we get $x = 1 \ m$.
Now,find the corresponding height $y$ at $x = 1 \ m$:
$y = \frac{x^2}{4} = \frac{(1)^2}{4} = 0.25 \ m$.
Since $1 \ m = 100 \ cm$,the height is $0.25 \times 100 = 25 \ cm$.

(A) For a smooth inclined plane:
The acceleration is $a_1 = g \sin 30^{\circ} = \frac{g}{2}$.
Using the equation of motion $S = ut + \frac{1}{2}at^2$,with $u = 0$:
$S = \frac{1}{2} \left(\frac{g}{2}\right) T^2 = \frac{g T^2}{4} \quad \dots (i)$
For a rough inclined plane:
The acceleration is $a_2 = g \sin 30^{\circ} - \mu g \cos 30^{\circ} = g(\sin 30^{\circ} - \mu \cos 30^{\circ}) = g\left(\frac{1}{2} - \mu \frac{\sqrt{3}}{2}\right) = \frac{g}{2}(1 - \sqrt{3}\mu)$.
Using the same distance $S$ and time $\alpha T$:
$S = \frac{1}{2} \left[\frac{g}{2}(1 - \sqrt{3}\mu)\right] (\alpha T)^2 = \frac{g}{4}(1 - \sqrt{3}\mu) \alpha^2 T^2 \quad \dots (ii)$
Equating $(i)$ and $(ii)$:
$\frac{g T^2}{4} = \frac{g}{4}(1 - \sqrt{3}\mu) \alpha^2 T^2$
$1 = (1 - \sqrt{3}\mu) \alpha^2$
$\frac{1}{\alpha^2} = 1 - \sqrt{3}\mu$
$\sqrt{3}\mu = 1 - \frac{1}{\alpha^2} = \frac{\alpha^2 - 1}{\alpha^2}$
$\mu = \frac{1}{\sqrt{3}} \left(\frac{\alpha^2 - 1}{\alpha^2}\right)$
Comparing this with the given expression $\frac{1}{\sqrt{x}}\left(\frac{\alpha^{2}-1}{\alpha^{2}}\right)$,we get $x = 3$.

(C) Let $t_a$ be the time of ascent and $t_d$ be the time of descent. Given $t_a = \frac{1}{2} t_d$,which implies $t_d = 2 t_a$.
Since the distance $s$ covered is the same,$s = \frac{1}{2} a_a t_a^2 = \frac{1}{2} a_d t_d^2$.
Substituting $t_d = 2 t_a$,we get $a_a t_a^2 = a_d (2 t_a)^2$,so $a_a = 4 a_d$.
The acceleration during ascent is $a_a = g \sin \theta + \mu g \cos \theta = g \sin 30^{\circ} + \mu g \cos 30^{\circ} = \frac{g}{2} + \frac{\sqrt{3}}{2} \mu g$.
The acceleration during descent is $a_d = g \sin \theta - \mu g \cos \theta = g \sin 30^{\circ} - \mu g \cos 30^{\circ} = \frac{g}{2} - \frac{\sqrt{3}}{2} \mu g$.
Substituting these into $a_a = 4 a_d$: $\frac{g}{2} + \frac{\sqrt{3}}{2} \mu g = 4 (\frac{g}{2} - \frac{\sqrt{3}}{2} \mu g)$.
Dividing by $g$: $\frac{1}{2} + \frac{\sqrt{3}}{2} \mu = 2 - 2\sqrt{3} \mu$.
$\frac{5\sqrt{3}}{2} \mu = \frac{3}{2} \implies \mu = \frac{3}{5\sqrt{3}} = \frac{\sqrt{3}}{5}$.
Comparing $\mu = \frac{\sqrt{3}}{5}$ with $\frac{\sqrt{x}}{5}$,we get $x = 3$.

(A) Since the block slides down with constant velocity,the net force acting on it is zero.
The forces acting on the block are the gravitational force $(Mg)$,the normal force $(N)$,and the kinetic friction force $(f)$.
Resolving the gravitational force into components,we have:
$N = Mg \cos \theta$ (perpendicular to the plane)
$f = Mg \sin \theta$ (parallel to the plane)
The contact force $(R)$ is the resultant of the normal force $(N)$ and the friction force $(f)$:
$R = \sqrt{N^2 + f^2}$
Substituting the values of $N$ and $f$:
$R = \sqrt{(Mg \cos \theta)^2 + (Mg \sin \theta)^2}$
$R = \sqrt{M^2g^2(\cos^2 \theta + \sin^2 \theta)}$
Since $\sin^2 \theta + \cos^2 \theta = 1$,we get:
$R = \sqrt{M^2g^2(1)}$
$R = Mg$