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

(C) Let the total length of the rope be $l$ and its total mass be $M$. The mass per unit length is $\lambda = \frac{M}{l}$.
Let $l_1$ be the length of the rope overhanging the table. The length of the rope on the table is $(l - l_1)$.
The mass of the overhanging part is $m_1 = \lambda l_1 = \frac{M}{l} l_1$. The weight of this part acts as the force pulling the rope down: $F_g = m_1 g = \frac{M g l_1}{l}$.
The mass of the part on the table is $m_2 = \lambda (l - l_1) = \frac{M}{l} (l - l_1)$. The normal force on this part is $N = m_2 g = \frac{M g (l - l_1)}{l}$.
The maximum static frictional force is $f_{max} = \mu N = \mu \frac{M g (l - l_1)}{l}$.
For the rope to be in equilibrium without sliding,the pulling force must be equal to the maximum frictional force: $F_g = f_{max}$.
$\frac{M g l_1}{l} = \mu \frac{M g (l - l_1)}{l}$.
$l_1 = \mu (l - l_1)$.
$l_1 = \mu l - \mu l_1$.
$l_1 (1 + \mu) = \mu l$.
$l_1 = \frac{\mu l}{1 + \mu}$.

(A) Let the total length of the chain be $L$ and its total mass be $M$. The mass per unit length is $\lambda = M/L$.
Let $x$ be the length of the chain hanging over the edge. Then the length of the chain on the table is $(L - x)$.
The mass of the hanging part is $m_h = \lambda x$ and the mass of the part on the table is $m_t = \lambda (L - x)$.
The force pulling the chain down is the weight of the hanging part: $F_g = m_h g = \lambda x g$.
The maximum static frictional force acting on the part on the table is $f_{max} = \mu N = \mu m_t g = \mu \lambda (L - x) g$.
For the chain to be on the verge of sliding,the pulling force must equal the maximum frictional force: $\lambda x g = \mu \lambda (L - x) g$.
Dividing both sides by $\lambda g$,we get $x = \mu (L - x)$.
Substituting $\mu = 0.25$: $x = 0.25(L - x) \implies x = 0.25L - 0.25x \implies 1.25x = 0.25L$.
Thus,the fraction $x/L = 0.25 / 1.25 = 1/5 = 0.20$.
Converting to percentage,the fraction is $20\%$.

(C) Let the total length of the chain be $L$ and the mass per unit length be $\lambda$.
The length of the chain hanging from the table is $l$.
The length of the chain lying on the table is $(L - l)$.
The weight of the hanging part,which acts as the pulling force,is $F_g = (\lambda l)g$.
The normal force acting on the part of the chain on the table is $N = (\lambda(L - l))g$.
For the chain to be in equilibrium,the limiting friction must balance the weight of the hanging part: $f_{max} = \mu N = F_g$.
Substituting the values: $\mu (\lambda(L - l))g = (\lambda l)g$.
Solving for $\mu$: $\mu = \frac{l}{L - l}$.

(B) When a body is placed on an inclined plane,the angle at which it just begins to slide is called the angle of repose $(\theta)$.
At this angle,the component of gravitational force acting down the plane is equal to the maximum static frictional force.
$mg \sin \theta = \mu_s mg \cos \theta$
Therefore,the coefficient of static friction $\mu_s = \tan \theta$.
Given that the angle of inclination $\theta = 60^\circ$,
$\mu_s = \tan 60^\circ = \sqrt{3} \approx 1.732$.
Thus,the correct option is $B$.

(D) Let the total length of the chain be $L$ and its total mass be $M$. The mass per unit length is $\lambda = M/L$.
When $1/3$ of the length hangs over the edge,the length of the hanging part is $L_h = L/3$ and the length on the table is $L_t = 2L/3$.
The mass of the hanging part is $m_h = \lambda (L/3) = M/3$.
The mass of the part on the table is $m_t = \lambda (2L/3) = 2M/3$.
The force pulling the chain down is the weight of the hanging part: $F_g = m_h g = (M/3)g$.
The normal force on the part of the chain on the table is $N = m_t g = (2M/3)g$.
The limiting friction force is $f_s = \mu_s N = \mu_s (2M/3)g$.
At the point of sliding,the pulling force equals the limiting friction force: $F_g = f_s$.
$(M/3)g = \mu_s (2M/3)g$.
Solving for $\mu_s$: $\mu_s = (M/3) / (2M/3) = 1/2$.

(B) When a body is at rest on an inclined plane,the force of friction acting on it is static friction.
Static friction is a self-adjusting force that balances the component of the gravitational force acting down the plane $(mg \sin \theta)$.
Since the body is not moving,the static friction is less than or equal to the limiting friction $(f_s \leq \mu R)$.
Therefore,in the general case where the body is not on the verge of sliding,the force of friction is less than $\mu R$.

(A) For a body of mass $m$ on an inclined plane with angle of inclination $\theta$ and coefficient of friction $\mu$,the forces acting along the plane are the component of gravity $mg \sin \theta$ acting downwards and the frictional force $f = \mu N$ acting upwards.
The normal force $N$ is given by $N = mg \cos \theta$.
The net force $F_{\text{net}}$ acting down the plane is $F_{\text{net}} = mg \sin \theta - f = mg \sin \theta - \mu mg \cos \theta$.
Using Newton's second law,$F_{\text{net}} = ma$,we get $ma = mg(\sin \theta - \mu \cos \theta)$.
Therefore,the acceleration $a$ is $g(\sin \theta - \mu \cos \theta)$.

(C) When a block is placed on an inclined plane,the forces acting on it are the gravitational force $(mg)$,the normal force $(N)$,and the static frictional force $(f_s)$.
The component of weight acting down the plane is $mg \sin \alpha$ and the component perpendicular to the plane is $mg \cos \alpha$.
For the block to be at rest,the frictional force $f_s = mg \sin \alpha$ and the normal force $N = mg \cos \alpha$.
The block starts slipping when the angle of inclination reaches the angle of repose,$\theta$. At this point,the static friction reaches its maximum value,$f_{s,max} = \mu_s N$.
Equating the forces at the point of slipping: $mg \sin \theta = \mu_s (mg \cos \theta)$.
Dividing both sides by $mg \cos \theta$,we get $\mu_s = \frac{\sin \theta}{\cos \theta} = \tan \theta$.
Thus,the coefficient of static friction is $\tan \theta$.

(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 $t_r = \sqrt{\frac{2s}{g(\sin \theta - \mu \cos \theta)}}$.
Given $t_r = n t_s$,so $t_r^2 = n^2 t_s^2$.
$\frac{2s}{g(\sin \theta - \mu \cos \theta)} = n^2 \frac{2s}{g \sin \theta}$.
$\sin \theta - \mu \cos \theta = \frac{\sin \theta}{n^2}$.
$\mu \cos \theta = \sin \theta \left( 1 - \frac{1}{n^2} \right)$.
$\mu = \tan \theta \left( 1 - \frac{1}{n^2} \right)$.
Since $\theta = 45^{\circ}$,$\tan 45^{\circ} = 1$,so $\mu = 1 - \frac{1}{n^2}$.

(A) Let $m$ be the mass of the body,$\theta$ be the angle of inclination,and $\mu$ be the coefficient of friction.
For moving the body up the plane,the applied force $F_{up}$ must overcome both the component of gravity $mg \sin \theta$ and the frictional force $\mu mg \cos \theta$.
Thus,$F_{up} = mg(\sin \theta + \mu \cos \theta)$.
To prevent the body from sliding down,the applied force $F_{dn}$ acts upwards,balancing the component of gravity $mg \sin \theta$ minus the frictional force $\mu mg \cos \theta$ (which acts upwards to oppose sliding).
Thus,$F_{dn} = mg(\sin \theta - \mu \cos \theta)$.
Given that $F_{up} = 2F_{dn}$,we have:
$mg(\sin \theta + \mu \cos \theta) = 2mg(\sin \theta - \mu \cos \theta)$.
Dividing by $mg$,we get $\sin \theta + \mu \cos \theta = 2\sin \theta - 2\mu \cos \theta$.
Rearranging the terms,we get $3\mu \cos \theta = \sin \theta$,which implies $\tan \theta = 3\mu$.
Given $\mu = 0.25$,$\tan \theta = 3 \times 0.25 = 0.75$.
Therefore,$\theta = \tan^{-1}(0.75) \approx 36.8^\circ$.

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

(D) The angle of repose is defined as the minimum angle of an inclined plane at which a body placed on it just begins to slide down.
If the angle of inclination $\theta$ is equal to the angle of repose $\alpha$,the body is on the verge of sliding.
If the angle of inclination $\theta$ is greater than the angle of repose $\alpha$,the body will slide down automatically due to gravity.
Since the box has to be pushed down,it implies that the box is not sliding on its own,which means the gravitational component pulling it down is less than the maximum static frictional force.
Therefore,the angle of inclination must be less than the angle of repose. Hence,the correct option is $D$.

(D) $1$. The component of the gravitational force acting down the plane is $mg \sin \theta = 102 \times 9.8 \times \sin 30^{\circ} = 102 \times 9.8 \times 0.5 = 499.8 \, N \approx 500 \, N$.
$2$. The applied force $P = 750 \, N$ acts up the plane.
$3$. The net external force along the plane is $F_{\text{net}} = P - mg \sin \theta = 750 - 500 = 250 \, N$.
$4$. The normal reaction is $R = mg \cos \theta = 102 \times 9.8 \times \cos 30^{\circ} = 102 \times 9.8 \times 0.866 \approx 865.6 \, N$.
$5$. The limiting static friction is $f_{l} = \mu_s R = 0.4 \times 865.6 = 346.24 \, N$.
$6$. Since the net external force $(250 \, N)$ is less than the limiting static friction $(346.24 \, N)$,the block remains in equilibrium.
$7$. According to the law of static friction,the actual frictional force adjusts itself to balance the net external force. Therefore,the frictional force is $250 \, N$ acting down the plane.

(C) The acceleration $a$ of a block sliding down an inclined plane with friction is given by the formula: $a = g(\sin \theta - \mu \cos \theta)$.
Given values are: $g = 10\,m/s^2$,$\theta = 60^\circ$,and $\mu = 0.25$.
Substituting these values into the formula:
$a = 10 \times (\sin 60^\circ - 0.25 \times \cos 60^\circ)$
$a = 10 \times (\frac{\sqrt{3}}{2} - 0.25 \times \frac{1}{2})$
$a = 10 \times (0.866 - 0.125)$
$a = 10 \times 0.741$
$a \approx 7.4\,m/s^2$.

(B) The frictional force $F_k$ acting on a body sliding down an inclined plane is given by the formula: $F_k = \mu N$,where $N$ is the normal reaction.
For an inclined plane,the normal reaction $N = mg \cos \theta$.
Given: mass $m = 100\, g = 0.1\, kg$,inclination $\theta = 30^\circ$,$\mu = 1.7$,and taking acceleration due to gravity $g = 10\, m/s^2$.
Substituting the values: $F_k = 1.7 \times (0.1\, kg) \times (10\, m/s^2) \times \cos 30^\circ$.
$F_k = 1.7 \times 1 \times \frac{\sqrt{3}}{2} = 1.7 \times \sqrt{3} \times \frac{1}{2}\,N$.

(A) Let $L$ be the length of the inclined plane and $\theta = 30^\circ$ be the angle of inclination.
For a frictionless plane,the acceleration is $a_1 = g \sin \theta$. The time taken is $t_1 = \sqrt{\frac{2L}{g \sin \theta}}$.
For a rough plane with coefficient of friction $\mu$,the acceleration is $a_2 = g(\sin \theta - \mu \cos \theta)$. The time taken is $t_2 = \sqrt{\frac{2L}{g(\sin \theta - \mu \cos \theta)}}$.
Given $t_2 = 2t_1$,we have $\frac{t_2}{t_1} = 2$,so $\frac{t_2^2}{t_1^2} = 4$.
Substituting the expressions,$\frac{\sin \theta}{\sin \theta - \mu \cos \theta} = 4$.
$\sin \theta = 4 \sin \theta - 4 \mu \cos \theta \implies 4 \mu \cos \theta = 3 \sin \theta$.
$\mu = \frac{3}{4} \tan \theta = \frac{3}{4} \tan 30^\circ = \frac{3}{4} \times \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{4}$.

(A) When an object begins to slide down an inclined plane,it is at the angle of repose.
At the angle of repose,the force of static friction is equal to the component of the gravitational force acting down the plane.
The component of weight acting down the plane is given by $mg \sin \theta$.
Given: mass $m = 2 \, kg$,angle $\theta = 45^\circ$,and acceleration due to gravity $g = 9.8 \, m/s^2$.
Therefore,the force of friction $f = mg \sin \theta = 2 \times 9.8 \times \sin 45^\circ = 19.6 \sin 45^\circ$.

(D) 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 $a_1 = g \sin \theta$. Starting from rest $(u=0)$,the velocity $v$ at the midpoint is given by:
$v^2 = u^2 + 2 a_1 (l/2) = 0 + 2(g \sin \theta)(l/2) = gl \sin \theta$.
For the lower half (rough):
The acceleration is $a_2 = g(\sin \theta - \mu \cos \theta)$. The body starts with velocity $v$ and comes to rest $(v_f = 0)$ at the bottom after covering distance $l/2$:
$v_f^2 = v^2 + 2 a_2 (l/2) = 0$.
Substituting $v^2 = gl \sin \theta$ and $a_2 = g(\sin \theta - \mu \cos \theta)$:
$gl \sin \theta + 2[g(\sin \theta - \mu \cos \theta)](l/2) = 0$.
$gl \sin \theta + gl(\sin \theta - \mu \cos \theta) = 0$.
$2 \sin \theta - \mu \cos \theta = 0$.
$\mu \cos \theta = 2 \sin \theta$.
$\mu = 2 \tan \theta$.

(C) The resultant downward force along the incline is given by $F_{net} = mg(\sin \theta - \mu \cos \theta)$.
The normal reaction force is $N = mg \cos \theta$.
According to the problem,the normal reaction is twice the resultant downward force:
$mg \cos \theta = 2 \times mg(\sin \theta - \mu \cos \theta)$.
Dividing both sides by $mg$:
$\cos \theta = 2(\sin \theta - 0.5 \cos \theta)$.
Expanding the equation:
$\cos \theta = 2 \sin \theta - 1.0 \cos \theta$.
Rearranging the terms:
$2 \cos \theta = 2 \sin \theta$.
Dividing by $2 \cos \theta$:
$1 = \tan \theta$.
Therefore,$\theta = \tan^{-1}(1) = 45^o$.

(B) The force required to pull a body of mass $m$ up an inclined plane of angle $\theta$ with coefficient of friction $\mu$ is given by the formula:
$F = mg(\sin \theta + \mu \cos \theta)$
Given:
$m = 10\, kg$
$\theta = 30^\circ$
$\mu = 0.5$
$g = 9.8\, m/s^2$
Substituting the values:
$F = 10 \times 9.8 \times (\sin 30^\circ + 0.5 \cos 30^\circ)$
$F = 98 \times (0.5 + 0.5 \times 0.866)$
$F = 98 \times (0.5 + 0.433)$
$F = 98 \times 0.933 = 91.434\, N$
Rounding to the nearest provided option,the correct value is $91.4\, N$.

(C) The work done against friction $(W_f)$ is given by the product of the frictional force $(f_k)$ and the displacement $(S)$.
$f_k = \mu_k N = \mu_k mg \cos \theta$
Given: mass $m = 1 \,kg$,coefficient of kinetic friction $\mu_k = 0.5$,inclination $\theta = 60^\circ$,and length $S = 1 \,m$.
Using $g = 9.8 \,m/s^2$:
$W_f = f_k \times S = (\mu_k mg \cos \theta) \times S$
$W_f = 0.5 \times 1 \times 9.8 \times \cos(60^\circ) \times 1$
Since $\cos(60^\circ) = 0.5$:
$W_f = 0.5 \times 9.8 \times 0.5 \times 1 = 2.45 \,J$.

(D) The block is on the verge of sliding down the inclined plane. At this point,the component of the gravitational force acting down the plane is balanced by the maximum static friction force.
The force acting down the plane is $F = mg \sin \theta$.
Given $m = 10\, kg$,$\theta = 30^\circ$,and $g \approx 9.8\, m/s^2$.
The force of static friction $f_s = mg \sin 30^\circ = 10 \times 9.8 \times 0.5 = 49\, N$.
Since $1\, kg\, wt = 9.8\, N$,the force in $kg\, wt$ is $f_s = \frac{49}{9.8} = 5\, kg\, wt$.

(A) The angle of repose is given by $\alpha = \tan^{-1}(\mu) = \tan^{-1}(0.8) \approx 38.6^{\circ}$.
Since the angle of the inclined plane $\theta = 30^{\circ}$ is less than the angle of repose $\alpha$,the block remains at rest.
For a block at rest on an inclined plane,the static frictional force $f_s$ balances the component of gravity acting down the plane:
$f_s = mg \sin \theta$
Given $f_s = 10 \, N$,$g = 10 \, m/s^2$,and $\theta = 30^{\circ}$:
$10 = m \times 10 \times \sin(30^{\circ})$
$10 = m \times 10 \times 0.5$
$10 = 5m$
$m = 2 \, kg$.

(A) For a smooth inclined plane,the acceleration is $a_1 = g \sin \theta$. The distance $s$ is given by $s = \frac{1}{2} a_1 t^2 = \frac{1}{2} (g \sin \theta) t^2$.
For a rough inclined plane,the acceleration is $a_2 = g(\sin \theta - \mu \cos \theta)$. The distance $s$ is $s = \frac{1}{2} a_2 (2t)^2 = \frac{1}{2} g(\sin \theta - \mu \cos \theta) (4t^2)$.
Equating the two expressions for $s$: $\frac{1}{2} g \sin \theta t^2 = 2 g(\sin \theta - \mu \cos \theta) t^2$.
Dividing by $g t^2$: $\frac{1}{2} \sin \theta = 2(\sin \theta - \mu \cos \theta)$.
$\frac{1}{4} \sin \theta = \sin \theta - \mu \cos \theta$.
$\mu \cos \theta = \sin \theta - \frac{1}{4} \sin \theta = \frac{3}{4} \sin \theta$.
Therefore,$\mu = \frac{3}{4} \tan \theta$.

(B) The forces acting on the block along the inclined plane are the component of gravity $mg \sin \theta$ acting downwards and the frictional force $f_k = \mu N = \mu mg \cos \theta$ acting upwards.
Applying Newton's second law along the incline: $mg \sin \theta - \mu mg \cos \theta = ma$.
Thus,the acceleration of the block is $a = g(\sin \theta - \mu \cos \theta)$.
Using the kinematic equation $v^2 = u^2 + 2as$,where initial velocity $u = 0$ and distance $s = l$:
$v^2 = 0 + 2 \cdot g(\sin \theta - \mu \cos \theta) \cdot l$.
Therefore,the velocity at the bottom is $v = \sqrt{2gl(\sin \theta - \mu \cos \theta)}$.

(B) The normal force $R$ exerted by the wall on the block is equal to the applied horizontal force,so $R = 5 \, N$.
The limiting friction $F_l$ is given by $F_l = \mu R = 0.5 \times 5 = 2.5 \, N$.
The weight of the block acting downwards is $W = mg = 0.1 \times 9.8 = 0.98 \, N$.
Since the downward force (weight) is less than the limiting friction $(0.98 \, N < 2.5 \, N)$,the block remains at rest.
For a block in equilibrium,the static frictional force $F_s$ must balance the downward weight. Therefore,$F_s = W = 0.98 \, N$.

(A) The component of the weight acting down the inclined plane is $F_{applied} = mg \sin \theta = 2 \times 9.8 \times \sin 30^\circ = 2 \times 9.8 \times 0.5 = 9.8 \, N$.
The limiting friction is given by $F_l = \mu_s N = \mu_s mg \cos \theta$.
$F_l = 0.7 \times 2 \times 9.8 \times \cos 30^\circ = 0.7 \times 19.6 \times \frac{\sqrt{3}}{2} \approx 11.88 \, N$.
Since the applied force $(9.8 \, N)$ is less than the limiting friction $(11.88 \, N)$,the block remains at rest.
Therefore,the static frictional force is equal to the component of the weight acting down the plane,which is $9.8 \, N$.

(C) Given: Mass $m = 2 \ kg$,length of incline $S = 8 \ m$,height $h = 1 \ m$,coefficient of friction $\mu = 0.2$,$g = 10 \ m/s^2$.
From the geometry,$\sin \theta = h/S = 1/8$.
Using $\sin^2 \theta + \cos^2 \theta = 1$,we get $\cos \theta = \sqrt{1 - (1/8)^2} = \sqrt{63/64} = \sqrt{63}/8 \approx 7.937/8 \approx 0.992$.
To move the body at a constant speed,the applied force $F$ must balance the gravitational component and the frictional force: $F = mg \sin \theta + \mu mg \cos \theta$.
The work done $W = F \times S = (mg \sin \theta + \mu mg \cos \theta) \times S$.
$W = mgS \sin \theta + \mu mgS \cos \theta = mgh + \mu mgS \cos \theta$.
$W = (2 \times 10 \times 1) + (0.2 \times 2 \times 10 \times 8 \times 0.992) = 20 + 31.74 = 51.74 \ J$.
Rounding to the nearest integer provided in the options,the correct answer is $51 \ J$.

(A) Given: Mass $m = 1000 \; kg$,speed $v = 36 \; km/h = 36 \times (5/18) = 10 \; m/s$,$\tan \theta = 1/2$,$\mu = 1/\sqrt{3}$.
From the triangle,$\sin \theta = 1/\sqrt{5}$ and $\cos \theta = 2/\sqrt{5}$.
The force required to move the engine up the incline at a constant speed is $F = mg \sin \theta + f_k = mg \sin \theta + \mu mg \cos \theta$.
$F = mg(\sin \theta + \mu \cos \theta) = 1000 \times 9.8 \times (1/\sqrt{5} + (1/\sqrt{3}) \times (2/\sqrt{5}))$.
$F = 9800 \times (1/\sqrt{5} + 2/\sqrt{15}) = 9800 \times (0.4472 + 0.5164) \approx 9800 \times 0.9636 \approx 9443 \; N$.
Power $P = F \times v = 9443 \times 10 = 94430 \; W = 94.4 \times 10^3 \; W$.

(A) When a block moves 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 kinetic frictional force.
Let $m$ be the mass,$\theta$ be the angle of inclination,and $\mu_k$ be the coefficient of kinetic friction.
The force down the plane is $mg \sin \theta$.
The frictional force is $f_k = \mu_k N = \mu_k mg \cos \theta$.
Equating the two: $mg \sin \theta = \mu_k mg \cos \theta$.
Therefore,$\mu_k = \tan \theta$.
Given $\theta = 30^\circ$,we have $\mu_k = \tan 30^\circ = \frac{1}{\sqrt{3}}$.

(D) Let the total length of the incline be $L$. The length of each half is $l = L/2$.
For the smooth upper half,the acceleration is $a_1 = g \sin \theta$. Starting from rest $(u=0)$,the velocity $v$ at the midpoint is given by:
$v^2 = u^2 + 2 a_1 l = 0 + 2(g \sin \theta)(L/2) = gL \sin \theta$.
For the rough lower half,the acceleration is $a_2 = g(\sin \theta - \mu \cos \theta)$. The block comes to rest at the bottom,so the final velocity is $0$. Using $v_f^2 = v^2 + 2 a_2 l$:
$0 = gL \sin \theta + 2[g(\sin \theta - \mu \cos \theta)](L/2)$.
$0 = gL \sin \theta + gL \sin \theta - gL \mu \cos \theta$.
$2 \sin \theta = \mu \cos \theta$.
Therefore,$\mu = 2 \tan \theta$.

(C) When a block moves up an inclined plane,the retardation $a$ is given by $a = g(\sin \theta + \mu \cos \theta)$.
Using the equation of motion $v = u - at$,where $v = 0$ (final velocity at rest),$u = 5 \, m/s$,and $t = 0.5 \, s$:
$0 = u - at \implies a = \frac{u}{t} = \frac{5}{0.5} = 10 \, m/s^2$.
Now,substitute $a = 10 \, m/s^2$,$g = 10 \, m/s^2$,and $\theta = 30^o$ into the retardation formula:
$10 = 10(\sin 30^o + \mu \cos 30^o)$.
$1 = 0.5 + \mu (\frac{\sqrt{3}}{2})$.
$0.5 = \mu (\frac{1.732}{2}) = \mu (0.866)$.
$\mu = \frac{0.5}{0.866} \approx 0.577$.
Rounding to the nearest provided option,the value is approximately $0.6$.

(A) The total work done $(W_{total})$ is the sum of the work done against gravity $(W_g)$ and the work done against friction $(W_f)$.
$W_{total} = W_g + W_f$
Given: $m = 2\,kg$,$h = 10\,m$,$g = 10\,m/s^2$,$W_{total} = 300\,J$.
The work done against gravity is $W_g = mgh = 2 \times 10 \times 10 = 200\,J$.
Therefore,the work done against friction is $W_f = W_{total} - W_g$.
$W_f = 300\,J - 200\,J = 100\,J$.

(A) For an object to remain at rest on an inclined surface,the angle of inclination $\alpha$ must be less than or equal to the angle of repose $\theta$,where $\tan \theta = \mu$.
Here,the surface is curved,and the angle $\alpha$ is measured with the vertical.
The forces acting on the ant are gravity ($mg$ downwards),normal force ($N$ perpendicular to the surface),and friction ($f$ along the tangent).
At the point of slipping,the friction is limiting,so $f = \mu N$.
Resolving forces along the tangent and normal to the surface:
$mg \sin \alpha = f = \mu N$
$mg \cos \alpha = N$
Dividing the two equations:
$\tan \alpha = \mu$
Given $\mu = 1/3$,we have $\tan \alpha = 1/3$.
Therefore,$\cot \alpha = 1/\tan \alpha = 3$.

(A) Let $L$ be the total length of the chain and $M$ be its total mass. The mass per unit length is $\lambda = M/L$.
Let $l$ be the length of the chain hanging off the table. The length of the chain on the table is $(L - l)$.
The weight of the hanging part is $W_h = (\lambda l)g$,which acts as the pulling force.
The normal force on the part of the chain on the table is $N = (\lambda(L - l))g$.
The limiting friction force is $f_{max} = \mu N = \mu \lambda(L - l)g$.
For the chain to be in equilibrium,the pulling force must equal the limiting friction force:
$\lambda l g = \mu \lambda(L - l)g$
$l = \mu(L - l)$
$l = \mu L - \mu l$
$l(1 + \mu) = \mu L$
$l/L = \mu / (1 + \mu)$
Given $\mu = 0.25$:
$l/L = 0.25 / (1 + 0.25) = 0.25 / 1.25 = 1/5 = 0.20$.
Thus,the percentage of the length is $0.20 \times 100 = 20\%$.

(A) For a frictionless inclined plane,the acceleration is $a_1 = g \sin \theta$. The time taken to cover distance $s$ is $t_1 = \sqrt{\frac{2s}{g \sin \theta}}$.
For a rough inclined plane,the acceleration is $a_2 = g(\sin \theta - \mu \cos \theta)$. The time taken is $t_2 = \sqrt{\frac{2s}{g(\sin \theta - \mu \cos \theta)}}$.
Given $t_2 = n t_1$,so $\frac{t_2}{t_1} = n$.
Thus,$\frac{\sin \theta}{\sin \theta - \mu \cos \theta} = n^2$.
Rearranging for $\mu$: $\mu = \tan \theta \left( 1 - \frac{1}{n^2} \right)$.
Substituting $\theta = 30^{\circ}$ and $n = 2$:
$\mu = \tan 30^{\circ} \left( 1 - \frac{1}{2^2} \right) = \frac{1}{\sqrt{3}} \left( 1 - \frac{1}{4} \right) = \frac{1}{\sqrt{3}} \left( \frac{3}{4} \right) = \frac{\sqrt{3}}{4}$.

(D) The child is sliding down the rope,so the gravitational force acts downwards and the frictional force acts upwards.
Weight of the child,$W = mg = 25 \times 9.8 = 245$ $N$.
Frictional force,$f = 2$ $N$.
According to Newton's second law,the net force $F_{net} = W - f$.
$ma = mg - f$
$25 \times a = 245 - 2$
$25a = 243$
$a = \frac{243}{25} = 9.72$ $m/s^2$.
Therefore,the acceleration of the child is $9.72$ $m/s^2$.

(B) Let the length of both grooves be $L$. The acceleration of a particle sliding down a smooth groove inclined at an angle $\theta$ with the vertical is $a = g \cos \theta$.
For the vertical chord $AB$,the angle with the vertical is $\theta_1 = 0^\circ$. Thus,$a_{AB} = g \cos 0^\circ = g$.
The time taken is $t_{AB} = \sqrt{\frac{2L}{a_{AB}}} = \sqrt{\frac{2L}{g}}$.
For the chord $CD$,the angle with the vertical is $\theta_2 = 60^\circ$. Thus,$a_{CD} = g \cos 60^\circ = g/2$.
The time taken is $t_{CD} = \sqrt{\frac{2L}{a_{CD}}} = \sqrt{\frac{2L}{g/2}} = \sqrt{\frac{4L}{g}} = 2\sqrt{\frac{L}{g}}$.
Taking the ratio,$\frac{t_{AB}}{t_{CD}} = \frac{\sqrt{2L/g}}{2\sqrt{L/g}} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.

(B) 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 $N = mg \cos \theta$.
Case $1$: When the block is at rest (static friction),$f = mg \sin \theta$.
Then $F_c = \sqrt{(mg \cos \theta)^2 + (mg \sin \theta)^2} = \sqrt{m^2g^2(\cos^2 \theta + \sin^2 \theta)} = mg$.
This holds until the angle of repose $\theta_r$ is reached,where $\tan \theta_r = \mu$.
Case $2$: When the block starts sliding (kinetic friction),$f = \mu N = \mu mg \cos \theta$.
Then $F_c = \sqrt{N^2 + f^2} = \sqrt{(mg \cos \theta)^2 + (\mu mg \cos \theta)^2} = mg \cos \theta \sqrt{1 + \mu^2}$.
As $\theta$ increases from $0^o$ to $\theta_r$,$F_c$ remains constant at $mg$.
As $\theta$ increases from $\theta_r$ to $90^o$,$F_c = mg \cos \theta \sqrt{1 + \mu^2}$ decreases because $\cos \theta$ decreases.
Therefore,the contact force first remains constant and then decreases.

(A) From the Free Body Diagram $(FBD)$,both the gravitational component $(mg \sin 37^{\circ})$ and the frictional force $(f = \mu N = \mu mg \cos 37^{\circ})$ act downward along the inclined plane.
Therefore,the block experiences a constant deceleration $(a)$ while moving up the plane:
$ma = mg \sin 37^{\circ} + \mu mg \cos 37^{\circ}$
$a = g(\sin 37^{\circ} + \mu \cos 37^{\circ})$
Given $g = 10 \ m/s^2$,$\sin 37^{\circ} \approx 0.6$,$\cos 37^{\circ} \approx 0.8$,and $\mu = 0.5$:
$a = 10(0.6 + 0.5 \times 0.8) = 10(0.6 + 0.4) = 10 \ m/s^2$
Using the kinematic equation $v^2 = u^2 - 2aS$,where final velocity $v = 0$ and initial velocity $u = 5 \ m/s$:
$0 = (5)^2 - 2(10)S$
$20S = 25$
$S = 1.25 \ m$

(C) $1$. Analyze the forces on the $5 \, kg$ block: Since the system is in equilibrium,the tension $T$ in the string is equal to the weight of the $5 \, kg$ block. Thus,$T = m_2g = 5 \times 10 = 50 \, N$.
$2$. Analyze the forces on the $10 \, kg$ block along the inclined plane: The component of gravity acting down the plane is $m_1g \sin(37^\circ) = 10 \times 10 \times 0.6 = 60 \, N$.
$3$. Determine the direction of friction: The gravitational force component $(60 \, N)$ is greater than the tension $(50 \, N)$ pulling the block up the plane. Therefore,the block has a tendency to slide down the plane. To maintain equilibrium,the friction force must act up the plane.
$4$. Calculate the friction force: Let $f$ be the friction force. For equilibrium along the plane,$T + f = m_1g \sin(37^\circ)$. Substituting the values,$50 + f = 60$,which gives $f = 10 \, N$ up the plane.

(B) When the body hits the smooth inclined plane,the impulse acts perpendicular to the surface of the plane. Therefore,the component of velocity parallel to the inclined plane remains unchanged.
Let the initial velocity be $v = 3\, m/s$ directed vertically downward.
The angle of the inclined plane with the horizontal is $\theta = 30^{\circ}$.
The component of the initial velocity $v$ parallel to the inclined plane is $v_{\parallel} = v \sin 30^{\circ}$.
After the collision,the body moves horizontally with velocity $v_f$. The component of this final velocity $v_f$ parallel to the inclined plane is $v_{f\parallel} = v_f \cos 30^{\circ}$.
Since the surface is smooth,there is no force parallel to the plane,so the velocity component parallel to the plane is conserved:
$v \sin 30^{\circ} = v_f \cos 30^{\circ}$
$v_f = v \tan 30^{\circ}$
Substituting the values $v = 3\, m/s$ and $\tan 30^{\circ} = 1/\sqrt{3}$:
$v_f = 3 \times (1/\sqrt{3}) = \sqrt{3}\, m/s$.

(D) The acceleration of a block on an inclined plane is given by $a = g(\sin \theta - \mu \cos \theta)$.
Since the masses are equal,the block with the lower coefficient of friction $(\mu)$ will have a greater acceleration down the slope.
If $\mu_1 > \mu_2$,block $B$ (with $\mu_2$) tends to accelerate faster than block $A$ (with $\mu_1$). Since $B$ is behind $A$,they will remain in contact.
If $\mu_1 < \mu_2$,block $A$ tends to accelerate faster than block $B$. Since $A$ is in front of $B$,they will separate and slide down with different accelerations.
Therefore,both statements $(A)$ and $(B)$ are correct.

(C) For the system to accelerate,the driving force must overcome the limiting friction force acting on block $M$.
$1$. The limiting friction force on block $M$ is $f_{max} = \mu N = \mu M g$,where $N = M g$ is the normal force.
$2$. The driving force acting on the system is the weight of the hanging block $m$,which is $m g$.
$3$. The system will start to accelerate only when the driving force exceeds the limiting friction force:
$m g > \mu M g$
$4$. Dividing both sides by $g$,we get:
$m > \mu M$
Therefore,the system will accelerate only when $m > \mu M$.

(A) The energy lost due to friction $(E)$ as a block slides down an inclined plane of length $d$ is given by the work done against the frictional force.
The frictional force is $f = \mu_k N$, where $N = mg \cos \theta$ is the normal force.
Thus, the energy lost is $E = f \cdot d = \mu_k (mg \cos \theta) d$.
Analyzing the variables:
$1$. Increasing the angle of inclination $(\theta)$: As $\theta$ increases, $\cos \theta$ decreases, so the energy lost decreases.
$2$. Increasing the distance $(d)$: Energy lost increases linearly with $d$.
$3$. Increasing the acceleration due to gravity $(g)$: Energy lost increases linearly with $g$.
$4$. Increasing the mass $(m)$: Energy lost increases linearly with $m$.
Therefore, increasing the angle of inclination does not increase the energy lost; it decreases it.

(A) Consider the brick on an inclined plane. The center of mass of the brick is at its geometric center. For the brick to remain in equilibrium,the normal force from the inclined plane must act at a point such that the torque due to gravity about that point is zero.
Since the brick is on an incline,the normal force is shifted towards the lower edge (left side in the diagram) to counteract the torque produced by the gravitational force acting at the center of mass.
Because the normal force is distributed more towards the left half of the base,the left half of the brick exerts greater pressure on the inclined plane.