Kinetic Friction and Motion on Rough Horizontal Surface Questions in English

(D) $1$. First,calculate the acceleration of the entire rope. The net force acting on the rope is $F_{net} = F - f_k$,where $f_k = \mu N = \mu Mg = (1/2)Mg$.
$2$. Thus,$F_{net} = Mg - 0.5Mg = 0.5Mg$.
$3$. The acceleration $a$ is given by $a = F_{net} / M = 0.5Mg / M = g/2$.
$4$. Now,consider the rear half of the rope (mass $M/2$). The forces acting on this half are the tension $T$ at the midpoint pulling it forward and the kinetic friction $f_k'$ pulling it backward.
$5$. The friction on the rear half is $f_k' = \mu (M/2)g = (1/2)(M/2)g = Mg/4$.
$6$. Applying Newton's second law to the rear half: $T - f_k' = (M/2)a$.
$7$. Substituting the values: $T - Mg/4 = (M/2)(g/2) = Mg/4$.
$8$. Therefore,$T = Mg/4 + Mg/4 = Mg/2$.

(A) $1$. The crate is dropped vertically,so its initial horizontal velocity is $0$. The belt moves at a constant velocity $v = 1.20 \ m/s$.
$2$. The force of friction acting on the crate is $f = \mu mg = 0.400 \times 300 \times 9.8 = 1176 \ N$.
$3$. The acceleration of the crate is $a = f/m = \mu g = 0.400 \times 9.8 = 3.92 \ m/s^2$.
$4$. The time taken for the crate to reach the belt's speed is $t = v/a = 1.20 / 3.92 \approx 0.306 \ s$.
$5$. The distance moved by the belt during this time is $d_{belt} = v \times t = 1.20 \times (1.20 / 3.92) = 1.44 / 3.92 \approx 0.367 \ m$.
$6$. The motor must exert a force equal to the friction force to maintain constant speed,so $F_{motor} = f = 1176 \ N$.
$7$. The work done by the motor is $W = F_{motor} \times d_{belt} = 1176 \times (1.44 / 3.92) = 432 \ J$.

(D) First, analyze the forces acting on the block in the vertical direction to find the normal force $N$:
$N = W + \frac{W}{2} - \frac{W}{2} \sin(30^\circ) = W + \frac{W}{2} - \frac{W}{4} = \frac{5}{4} W$.
Next, analyze the forces in the horizontal direction. The driving force is $F_x = \frac{W}{2} \cos(30^\circ) = \frac{W}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3} W}{4}$.
The limiting friction force is $f_L = \mu N = \mu \cdot \frac{5}{4} W$.
The block will move if the driving force is greater than the limiting friction force, i.e., $F_x > f_L$.
$\frac{\sqrt{3} W}{4} > \mu \cdot \frac{5}{4} W \Rightarrow \mu < \frac{\sqrt{3}}{5}$.
Therefore, if $\mu < \frac{\sqrt{3}}{5}$, the block will move. If the block does not move, the work done by static friction is zero. Thus, option $D$ is correct.

(B) Given: Mass $m = 10\, kg$,initial velocity $u = 10\, m/s$,final velocity $v = 0\, m/s$,coefficient of friction $\mu = 0.2$,and acceleration due to gravity $g = 10\, m/s^2$.
According to Newton's second law,the frictional force is $f = \mu N = \mu mg$.
The retardation $a$ produced by this force is $a = \frac{f}{m} = \mu g$.
Substituting the values: $a = 0.2 \times 10 = 2\, m/s^2$.
Using the third equation of motion $v^2 = u^2 - 2as$:
$0^2 = 10^2 - 2 \times 2 \times s$
$0 = 100 - 4s$
$4s = 100$
$s = 25\, m$.
Therefore,the block will stop after travelling a distance of $25\, m$.

(C) The forces acting on the block are the applied force $F$,gravity $mg$,normal force $N$,and friction $f$.
Resolving the force $F$ into components:
Horizontal component: $F_x = F \cos 37^{\circ} = 5t \times \frac{4}{5} = 4t$
Vertical component: $F_y = F \sin 37^{\circ} = 5t \times \frac{3}{5} = 3t$
For vertical equilibrium:
$N + F_y = mg$
$N = mg - F_y = 70 - 3t$
The block starts to slide when the horizontal force $F_x$ equals the limiting friction $f_L = \mu N$:
$F_x = \mu N$
$4t = 1 \times (70 - 3t)$
$4t = 70 - 3t$
$7t = 70$
$t = 10\,s$

(D) Given: $M = \frac{10}{3} \, kg$,$F = 50 \, N$,$\mu = \frac{1}{3}$,$\theta = \sin^{-1}(\frac{3}{5})$.
From $\sin \theta = \frac{3}{5}$,we get $\cos \theta = \frac{4}{5}$.
The vertical forces on the block are the normal force $N$,the weight $Mg$,and the vertical component of the applied force $F \sin \theta$ (acting upwards).
$N + F \sin \theta = Mg \Rightarrow N = Mg - F \sin \theta$.
$N = (\frac{10}{3} \times 10) - (50 \times \frac{3}{5}) = \frac{100}{3} - 30 = \frac{100 - 90}{3} = \frac{10}{3} \, N$.
The horizontal forces are the horizontal component of the applied force $F \cos \theta$ and the friction force $f = \mu N$.
The equation of motion is $F \cos \theta - \mu N = Ma$.
$50 \times \frac{4}{5} - \frac{1}{3} \times \frac{10}{3} = \frac{10}{3} a$.
$40 - \frac{10}{9} = \frac{10}{3} a$.
$\frac{360 - 10}{9} = \frac{10}{3} a \Rightarrow \frac{350}{9} = \frac{10}{3} a$.
$a = \frac{350}{9} \times \frac{3}{10} = \frac{35}{3} \, ms^{-2}$.

(D) For a block moving at a uniform velocity,the net force acting on it must be zero.
Let $F$ be the applied force. The vertical forces are the normal force $N$,the vertical component of the applied force $F \sin \theta$,and the weight $mg$. Thus,$N + F \sin \theta = mg$,which gives $N = mg - F \sin \theta$.
The horizontal forces are the horizontal component of the applied force $F \cos \theta$ and the kinetic friction $f_k = \mu N$. For uniform velocity,$F \cos \theta = f_k = \mu(mg - F \sin \theta)$.
Solving for $F$: $F \cos \theta = \mu mg - \mu F \sin \theta \implies F(\cos \theta + \mu \sin \theta) = \mu mg \implies F = \frac{\mu mg}{\cos \theta + \mu \sin \theta}$.
The work done $W$ during displacement $d$ is $W = (F \cos \theta) \times d$. However,since the work done by the applied force is requested,$W = F \cdot d \cdot \cos \theta$ is not the standard interpretation here; usually,it refers to the work done by the applied force $F$ in the direction of displacement. Given the options,the work done by the applied force is $W = F \cdot d = \frac{\mu mgd}{\cos \theta + \mu \sin \theta}$.

(C) For the block to be in equilibrium,the net force acting on it must be zero.
$1$. The horizontal force pressing the block against the wall is $F_h = mg$. This force provides the normal reaction $N = mg$.
$2$. The maximum static friction force available is $f_{max} = \mu N = \mu (mg)$.
$3$. Considering the vertical forces,the weight of the block $mg$ acts downwards,and an upward force of $\frac{mg}{2}$ is applied. Let $f$ be the friction force acting on the block.
$4$. For equilibrium in the vertical direction: $f + \frac{mg}{2} = mg$,which gives $f = mg - \frac{mg}{2} = \frac{mg}{2}$.
$5$. For the block to remain in equilibrium,the required friction force must be less than or equal to the maximum static friction: $f \le f_{max}$.
$6$. Therefore,$\frac{mg}{2} \le \mu (mg)$.
$7$. Solving for $\mu$,we get $\mu \ge \frac{1}{2}$,or $\mu \ge 0.5$. The minimum coefficient of friction is $0.5$.

(C) The mass of the rod is $m = 1 \, kg$,so the gravitational force acting downwards is $W = mg = 1 \times 10 = 10 \, N$.
For the rod to remain in equilibrium,the total upward frictional force $F_f$ must balance the weight: $F_f = mg = 10 \, N$.
Each finger exerts a normal force $N = 12 \, N$. Since there are two fingers,the total maximum static friction available is $f_{s,max} = 2 \times \mu \times N = 2 \times 0.5 \times 12 = 12 \, N$.
Since the required frictional force $(10 \, N)$ is less than the maximum available static friction $(12 \, N)$,the rod does not slip.
Therefore,the actual frictional force acting on the rod is equal to the weight,which is $10 \, N$.

(B) Given: mass $m = 60 \ kg$,angle $\theta = 37^{\circ}$,coefficient of friction $\mu = 1/3$,$g = 10 \ m/s^2$.
Since the block moves at a constant speed,the net force is zero $(a = 0)$.
Resolving the force $F$ into horizontal and vertical components:
Horizontal direction: $F \cos 37^{\circ} = \mu N$ --- $(1)$
Vertical direction: $N + F \sin 37^{\circ} = mg$ --- $(2)$
From $(2)$,$N = mg - F \sin 37^{\circ}$.
Substitute $N$ into $(1)$:
$F \cos 37^{\circ} = \mu (mg - F \sin 37^{\circ})$
$F \cos 37^{\circ} = \mu mg - \mu F \sin 37^{\circ}$
$F (\cos 37^{\circ} + \mu \sin 37^{\circ}) = \mu mg$
$F = \frac{\mu mg}{\cos 37^{\circ} + \mu \sin 37^{\circ}}$
Using $\cos 37^{\circ} = 4/5$ and $\sin 37^{\circ} = 3/5$:
$F = \frac{(1/3) \times 60 \times 10}{(4/5) + (1/3) \times (3/5)} = \frac{200}{(4/5) + (1/5)} = \frac{200}{1} = 200 \ N$.

(A) Given:
Mass of the block,$m = 5\,kg$
Applied horizontal force,$F = 40\,N$
Coefficient of kinetic friction,$\mu_k = 0.4$
Acceleration due to gravity,$g = 10\,m/s^2$
Step $1$: Calculate the kinetic frictional force $(f_k)$:
The normal force on the horizontal surface is $N = mg = 5 \times 10 = 50\,N$.
The kinetic frictional force is given by $f_k = \mu_k N = 0.4 \times 50 = 20\,N$.
Step $2$: Calculate the net force $(F_{net})$:
The net force acting on the block is $F_{net} = F - f_k = 40\,N - 20\,N = 20\,N$.
Step $3$: Calculate the acceleration $(a)$:
Using Newton's second law,$F_{net} = ma$.
$a = \frac{F_{net}}{m} = \frac{20\,N}{5\,kg} = 4\,m/s^2$.
Therefore,the acceleration of the block is $4\,m/s^2$.

(C) Since the block is moving with a constant speed,the net force acting on it is zero.
Therefore,the applied horizontal force must be equal to the force of kinetic friction.
$F_{ext} = F_{k}$
We know that the kinetic friction force is given by $F_{k} = \mu_{k} N$,where $N$ is the normal force.
For a block on a horizontal surface,$N = mg$.
Thus,$F_{ext} = \mu_{k} mg$.
Given: $F_{ext} = 4\,N$,$m = 0.5\,kg$,and $g = 10\,m/s^2$.
Substituting the values: $4 = \mu_{k} \times 0.5 \times 10$.
$4 = \mu_{k} \times 5$.
$\mu_{k} = \frac{4}{5} = 0.8$.

(A) The initial kinetic energy of the car is $K = \frac{1}{2} mv^2.$
The force of friction acting on the car is $f = \mu N = \mu mg.$
According to the work-energy theorem,the work done by the frictional force is equal to the change in kinetic energy of the car.
$W = \Delta K$
$-f \cdot s = 0 - \frac{1}{2} mv^2$
$\mu mg s = \frac{1}{2} mv^2$
Solving for the stopping distance $s$:
$s = \frac{v^2}{2\mu g}.$

(B) The force $F = 75\,N$ is applied at an angle of $37^{\circ}$ to the horizontal.
Resolving the force into components:
Horizontal component $N = F \cos 37^{\circ} = 75 \times 0.8 = 60\,N$. This is the normal force exerted by the wall on the block.
Vertical component $F_y = F \sin 37^{\circ} = 75 \times 0.6 = 45\,N$ (acting upwards).
The weight of the block is $mg = 1 \times 10 = 10\,N$ (acting downwards).
The force of kinetic friction is $f_k = \mu N = 0.25 \times 60 = 15\,N$. Since the vertical component of the applied force $(45\,N)$ is greater than the weight $(10\,N)$,the block moves upwards.
The friction force acts downwards to oppose the motion.
Applying Newton's second law in the vertical direction:
$F_y - mg - f_k = ma$
$45 - 10 - 15 = 1 \times a$
$20 = a$
Therefore,the acceleration $a = 20\,m/s^2$.

(A) When a block moves on a rough horizontal surface,the only horizontal force acting on it is the kinetic frictional force,$f_k = \mu_k N$.
Since the surface is horizontal,the normal force $N = mg$.
Therefore,the frictional force is $f_k = \mu_k mg$.
According to Newton's second law,the magnitude of acceleration $a$ is given by $F = ma$,so $ma = \mu_k mg$.
This simplifies to $a = \mu_k g$.
Given $\mu_k = 0.2$ and taking $g = 10\,m/s^2$,we get:
$a = 0.2 \times 10 = 2\,m/s^2$.

(A) The maximum static friction force is given by $f_{max} = \mu mg$.
Given $m = 4\,kg$,$g = 10\,m/s^2$,and $\mu = 0.8$,we have $f_{max} = 0.8 \times 4 \times 10 = 32\,N$.
The applied force at $t = 2\,s$ is $F = kt^2 = 2 \times (2)^2 = 8\,N$.
Since the applied force $F = 8\,N$ is less than the maximum static friction $f_{max} = 32\,N$,the block remains at rest.
According to the laws of static friction,the frictional force equals the applied force when the object is in equilibrium.
Therefore,the force of friction is $8\,N$.

(C) The frictional force acting on a body moving on a rough surface is kinetic friction.
Kinetic friction is given by the formula $f_k = \mu_k N$,where $\mu_k$ is the coefficient of kinetic friction and $N$ is the normal force.
Since the surface is level and the body is not accelerating vertically,the normal force $N$ is equal to the weight of the body $(mg)$.
Kinetic friction is independent of the velocity of the body as long as the surface properties and the normal force remain unchanged.
Therefore,even if the velocity increases from $2 \, m/s$ to $4 \, m/s$,the frictional force remains constant at $10 \, N$.

(A) The block has a mass $m=1 \, kg$. The forces acting on it are gravity ($mg=10 \, N$ downwards),the normal force ($N$ upwards),and the applied force $\vec{F}=\hat{i}+4 \hat{j}$.
Resolving forces in the vertical $(y)$ direction:
$N + F_y = mg$
$N + 4 = 10$
$N = 6 \, N$
The maximum static friction force is given by $f_{\max} = \mu N = 0.3 \times 6 = 1.8 \, N$.
The horizontal component of the applied force is $F_x = 1 \, N$.
Since the required horizontal force to move the block $(1 \, N)$ is less than the maximum static friction $(1.8 \, N)$,the block remains at rest.
Therefore,the static friction force will exactly balance the applied horizontal force to keep the block in equilibrium.
Thus,the friction force is $f = -F_x \hat{i} = -1 \hat{i} = -\hat{i}$.

(A) $1$. Resolve the applied force $F$ into horizontal and vertical components: $F \cos \phi$ (horizontal) and $F \sin \phi$ (vertical).
$2$. The vertical forces acting on the block are the normal reaction $R$ (upwards),the vertical component of the applied force $F \sin \phi$ (upwards),and the weight $Mg$ (downwards). For vertical equilibrium: $R + F \sin \phi = Mg$,so $R = Mg - F \sin \phi$.
$3$. The horizontal force causing motion is $F \cos \phi$. The kinetic friction force $f$ opposing the motion is given by $f = \mu R = \mu(Mg - F \sin \phi)$.
$4$. Applying Newton's second law in the horizontal direction: $F \cos \phi - f = Ma$.
$5$. Substituting the expression for $f$: $F \cos \phi - \mu(Mg - F \sin \phi) = Ma$.
$6$. Solving for acceleration $a$: $Ma = F \cos \phi - \mu Mg + \mu F \sin \phi$.
$7$. Dividing by $M$: $a = \frac{F}{M}(\cos \phi + \mu \sin \phi) - \mu g$.

(D) The forces acting on the block are its weight $mg$ downwards,the normal reaction $R$ upwards,and the applied force $F = mg$ at an angle $\theta$ with the vertical.
Resolving the force $F$ into components:
Horizontal component $= F \sin \theta = mg \sin \theta$
Vertical component $= F \cos \theta = mg \cos \theta$
For vertical equilibrium:
$R + F \cos \theta = mg$
$R = mg - mg \cos \theta = mg(1 - \cos \theta)$
The frictional force $f_r$ is given by:
$f_r = \mu R = \mu mg(1 - \cos \theta)$
The block will be pulled if the horizontal component of the force is greater than or equal to the frictional force:
$mg \sin \theta \geq \mu mg(1 - \cos \theta)$
$\sin \theta \geq \mu(1 - \cos \theta)$
Using trigonometric identities $\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$ and $1 - \cos \theta = 2 \sin^2 \frac{\theta}{2}$:
$2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \geq \mu (2 \sin^2 \frac{\theta}{2})$
$\cos \frac{\theta}{2} \geq \mu \sin \frac{\theta}{2}$
$\cot \frac{\theta}{2} \geq \mu$

(A) $1$. Resolve the applied force $F$ into two components: horizontal component $F \cos \theta$ and vertical component $F \sin \theta$.
$2$. The vertical forces acting on the block are the normal force $N$ (upward),the vertical component of the applied force $F \sin \theta$ (upward),and the weight $mg$ (downward). Since there is no vertical motion,$N + F \sin \theta = mg$,which gives $N = mg - F \sin \theta$.
$3$. The frictional force $f_r$ is given by $f_r = \mu N = \mu(mg - F \sin \theta)$.
$4$. The net horizontal force acting on the block is $F \cos \theta - f_r$. According to Newton's second law,$F \cos \theta - f_r = ma$.
$5$. Substituting the value of $f_r$: $F \cos \theta - \mu(mg - F \sin \theta) = ma$.
$6$. Solving for acceleration $a$: $a = \frac{F \cos \theta - \mu mg + \mu F \sin \theta}{m} = \frac{F}{m}(\cos \theta + \mu \sin \theta) - \mu g$.

(B) For the system to be in equilibrium:
For the block of mass $M_1$ with additional mass $m$ on it:
The normal force $N = (M_1 + m)g$.
The limiting frictional force $f = \mu N = \mu(M_1 + m)g$.
For equilibrium,the tension $T$ in the string must be equal to the frictional force:
$T = \mu(M_1 + m)g$.
For the block of mass $M_2$ hanging vertically:
For equilibrium,the tension $T$ must be equal to the weight of the block:
$T = M_2 g$.
Equating the two expressions for tension:
$\mu(M_1 + m)g = M_2 g$.
Dividing both sides by $\mu g$:
$M_1 + m = \frac{M_2}{\mu}$.
Therefore,the additional mass $m$ required is:
$m = \frac{M_2}{\mu} - M_1$.

(C) The force of kinetic friction $f$ acting on the body is given by $f = \mu R$.
To move the body slowly (without acceleration),an external force $P$ must be applied such that it exactly balances the frictional force.
Therefore,$P = f = \mu R$.
The work done $W$ against friction in moving the body by a distance $d$ is given by the product of the applied force and the displacement:
$W = P \times d = (\mu R) \times d = \mu Rd$.

(A) The box is dropped on the moving belt,so its initial velocity relative to the belt is $u_{rel} = 2\, ms^{-1}$.
The force of friction acting on the box is $f = \mu mg$.
According to Newton's second law,the acceleration of the box relative to the belt is $a = \frac{f}{m} = \mu g$.
Substituting the given values,$a = 0.5 \times 10 = 5\, ms^{-2}$.
We use the kinematic equation $v^2 - u^2 = 2aS$ to find the distance $S$ covered by the box relative to the belt until it stops sliding.
Here,the final relative velocity $v = 0\, ms^{-1}$ and the initial relative velocity $u = 2\, ms^{-1}$.
Substituting these into the equation: $0^2 - 2^2 = 2(-5)S$.
$-4 = -10S$.
$S = \frac{4}{10} = 0.4\, m$.

(C) The frictional force acting on the body is $f = \mu N = \mu mg$.
According to Newton's second law,the retardation $a$ is given by $ma = f = \mu mg$,which simplifies to $a = \mu g$.
Using the third equation of motion,$v_f^2 - v_i^2 = 2as$,where $v_f = 0$ (final velocity) and $v_i = v$ (initial velocity).
Substituting the values: $0^2 - v^2 = 2(-\mu g)s$.
This gives $-v^2 = -2\mu gs$.
Therefore,the initial velocity is $v = \sqrt{2\mu gs}$.

(C) Let $m_1 = 40\, kg$ be the mass on the table and $m_2 = 10\, kg$ be the hanging mass.
The limiting friction force acting on the $40\, kg$ block is $f_L = \mu m_1 g = 0.1 \times 40 \times 10 = 40\, N$.
The driving force due to the hanging mass is $F_d = m_2 g = 10 \times 10 = 100\, N$.
Since the driving force $(100\, N)$ is greater than the limiting friction $(40\, N)$,the system will accelerate.
Applying Newton's second law to the system: $m_2 g - f_L = (m_1 + m_2) a$.
$100 - 40 = (40 + 10) a$.
$60 = 50 a$.
$a = \frac{60}{50} = 1.2\, m/s^2$.

(A) Given: Initial velocity $u = 6\,m/s$,final velocity $v = 0\,m/s$,distance $s = 9\,m$,and acceleration due to gravity $g = 10\,m/s^2$.
Using the work-energy theorem,the work done by friction equals the change in kinetic energy: $W = \Delta K$.
$-f_k \cdot s = 0 - \frac{1}{2} m u^2$.
Since $f_k = \mu_k N = \mu_k m g$,we have $-\mu_k m g s = -\frac{1}{2} m u^2$.
Simplifying,$\mu_k = \frac{u^2}{2 g s}$.
Substituting the values: $\mu_k = \frac{6^2}{2 \times 10 \times 9} = \frac{36}{180} = 0.2$.

(D) Let $C$ be the center of mass of the block. The height of the center of mass from the base is $h/2$.
For the block to topple,the torque due to the frictional force $f$ about the edge of the base must be greater than the torque due to the normal force $N$ about the same edge.
Taking torque about the front edge of the base:
$\tau_{f} = f \times h$
$\tau_{N} = N \times \frac{a}{2}$
For toppling,$\tau_{f} > \tau_{N} \implies f \times h > N \times \frac{a}{2}$.
Since the block is moving,$f = \mu N$.
Substituting this,$\mu N h > N \frac{a}{2}$.
$\mu h > \frac{a}{2} \implies \mu > \frac{a}{2h}$.
Thus,the block will topple if $\mu > \frac{a}{2h}$.

(B) The block is moving to the right. The force $F = 20\sqrt{2}\, N$ acts at an angle of $45^{\circ}$ downwards, pushing the block into the floor.
Vertical forces: The normal reaction $N$ is given by $N = mg + F \sin 45^{\circ} = 5 \times 10 + 20\sqrt{2} \times (1/\sqrt{2}) = 50 + 20 = 70\, N$.
Frictional force: $f_k = \mu N = 0.5 \times 70 = 35\, N$.
Horizontal forces: The horizontal component of the applied force is $F_x = F \cos 45^{\circ} = 20\sqrt{2} \times (1/\sqrt{2}) = 20\, N$ (acting to the left).
The total retarding force is $F_{net} = f_k + F_x = 35 + 20 = 55\, N$.
Retardation $a = F_{net} / m = 55 / 5 = 11\, m/s^2$.
Using $v = u - at$, where $u = 33\, m/s$, $a = 11\, m/s^2$, and $t = 3\, s$:
$v = 33 - (11 \times 3) = 33 - 33 = 0\, m/s$.

(B) Given: Mass of the block $m = 5 \ kg$,Applied force $F = 1000 \ N$,Coefficient of friction $\mu = 0.1$,Acceleration due to gravity $g = 10 \ m/s^2$.
$1$. The block is pressed against the wall by a horizontal force $F = 1000 \ N$. This force acts as the normal force $N$ exerted by the wall on the block,so $N = F = 1000 \ N$.
$2$. The maximum static frictional force $f_{max}$ that can act is given by $f_{max} = \mu N = 0.1 \times 1000 = 100 \ N$.
$3$. The gravitational force acting downwards on the block is $W = mg = 5 \times 10 = 50 \ N$.
$4$. Since the block is in equilibrium,the upward frictional force $f$ must balance the downward gravitational force $W$. Therefore,$f = W = 50 \ N$.
$5$. Since $f = 50 \ N$ is less than $f_{max} = 100 \ N$,the block will remain in equilibrium. Thus,the frictional force is $50 \ N$.

(A) The frictional force acting on the body of mass $m$ is $f = \mu N = \mu mg$.
Since the platform is massive $(M \gg m)$,its velocity remains constant at $V_p = 4 \, m/s$.
The acceleration of the body $m$ is $a = \frac{f}{m} = \frac{\mu mg}{m} = \mu g$.
Substituting the values,$a = 0.2 \times 10 = 2 \, m/s^2$.
The body starts from rest $(u = 0)$ and accelerates until its velocity reaches the velocity of the platform $(v = V_p = 4 \, m/s)$.
Using the equation $v = u + at$,we get $4 = 0 + 2t$,which gives $t = 2 \, s$.
The distance traveled by the body in the ground frame is $S_b = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \times 2 \times (2)^2 = 4 \, m$.
The distance traveled by the platform in the same time is $S_p = V_p \times t = 4 \times 2 = 8 \, m$.
The sliding distance of the body relative to the platform is $S_{rel} = S_p - S_b = 8 - 4 = 4 \, m$.

(B) The frictional force $F$ provides the retarding acceleration $a$ to the car of mass $m$.
According to Newton's second law,$F = ma$.
The frictional force is also given by $F = \mu N$,where $\mu$ is the coefficient of friction and $N$ is the normal reaction.
For a car on a horizontal road,$N = mg$.
Equating the two expressions for $F$:
$ma = \mu mg$
$a = \mu g$
$\mu = \frac{a}{g}$
Given $a = 7.35\, ms^{-2}$ and taking $g = 9.8\, ms^{-2}$:
$\mu = \frac{7.35}{9.8} = 0.75$.

(C) The box remains stationary relative to the train floor due to the static frictional force acting on it.
Let $m$ be the mass of the box,$a$ be the acceleration of the train,and $\mu_s$ be the coefficient of static friction.
The force providing the acceleration to the box is the static friction $f_s$.
According to Newton's second law,$f_s = m a$.
For the box to remain stationary,the static friction must satisfy $f_s \leq \mu_s N$,where $N = m g$ is the normal force.
Thus,$m a \leq \mu_s m g$,which simplifies to $a \leq \mu_s g$.
The maximum acceleration $a_{\max}$ is given by $a_{\max} = \mu_s g$.
Substituting the given values,$a_{\max} = 0.15 \times 10 \ m/s^2 = 1.5 \ m/s^2$.

(N/A) When the magnitude of the force acting on an object lying on a surface exceeds the maximum static friction,the object will start moving in the direction of the external force. Hence,the value of the frictional force reduces to a value less than the maximum static frictional force.
The frictional force which opposes relative motion between surfaces in contact is called kinetic friction. It is denoted by $f_{k}$.
Laws of kinetic friction:
$(1)$ Kinetic friction does not depend on the area of contact between the surfaces.
$(2)$ Kinetic friction force does not depend on the relative velocity of the object in motion.
$(3)$ Kinetic friction force is proportional to the normal force.
$\therefore f_{k} \propto N$
$\therefore f_{k} = \mu_{k} N$
where $\mu_{k} = \text{coefficient of kinetic friction}$,$\mu_{k} = \frac{f_{k}}{N}$.
Coefficient of kinetic friction: The ratio of the kinetic friction force to the normal force is called the coefficient of kinetic friction.
Since $f_{s} > f_{k}$,it follows that $\mu_{s} > \mu_{k}$. Once relative motion starts,by Newton's second law of motion,the acceleration of the object is $a = \frac{F - f_{k}}{m}$.
If the object moves with constant velocity,then $F = f_{k}$.
If the external force applied is reduced to zero,then the acceleration of the object will be $-\frac{f_{k}}{m}$. Hence,it will stop after covering some distance.

(A) Rolling friction is the resistive force that opposes the motion of a body when it rolls over a surface without sliding. It is denoted by $f_{r}$.
Laws of rolling friction:
$(1)$ The magnitude of rolling friction is independent of the area of contact.
$(2)$ The magnitude of rolling friction is directly proportional to the normal force $(N)$: $f_{r} \propto N$,which implies $f_{r} = \mu_{r} N$.
Coefficient of rolling friction $(\mu_{r})$:
It is defined as the ratio of the rolling friction force to the normal force,given by $\mu_{r} = \frac{f_{r}}{N}$. It is a dimensionless quantity.
Explanation:
When a rigid body rolls over a surface,the surfaces in contact undergo momentary deformation. This deformation prevents the body from moving with constant velocity,as there is a resistive component of the contact force parallel to the surface. Rolling friction is significantly smaller than static or kinetic friction (typically $\frac{1}{100}$ to $\frac{1}{1000}$ times),which is why wheels are efficient. The relationship between coefficients is $\mu_{r} < \mu_{k} < \mu_{s}$.

(N/A) Kinetic friction is the resistive force that opposes the relative motion between two surfaces in contact when one surface is sliding over the other. It acts in the direction opposite to the direction of motion.
Rolling friction is the resistive force that opposes the motion of a body (like a wheel or sphere) when it rolls over a surface. It is generally much smaller than kinetic (sliding) friction because the area of contact is significantly reduced.

(D) The friction force always acts in the direction opposite to the relative motion or the tendency of motion of the object.
Since the instantaneous velocity represents the direction of motion,the friction force acts in the direction exactly opposite to the velocity vector.
Therefore,the angle between the instantaneous velocity and the friction force is $180^{\circ}$.

(A) The coefficient of friction between rubber and the road surface is significantly higher than that between steel and the road surface.
This higher coefficient of friction allows for better traction,which is essential for accelerating,braking,and turning the vehicle safely.
Steel tires would result in very low friction,causing the vehicle to skid easily.

(B) No,this statement is incorrect. When a man walks,he pushes the ground backward with his feet. According to Newton's third law,the ground exerts an equal and opposite force on his feet in the forward direction. This forward frictional force is what allows the man to move forward.

(B) $1$. Resolve the applied force $F$ into horizontal and vertical components: $F_{x} = F \cos \theta$ and $F_{y} = F \sin \theta$.
$2$. The vertical forces acting on the block are the normal force $N$ (upwards),the vertical component of the applied force $F \sin \theta$ (upwards),and the weight $mg$ (downwards). Since there is no vertical motion,the net vertical force is zero: $N + F \sin \theta = mg$,which gives $N = mg - F \sin \theta$.
$3$. The kinetic friction force $f_{k}$ is given by $f_{k} = \mu_{K} N = \mu_{K}(mg - F \sin \theta)$.
$4$. The net horizontal force acting on the block is the horizontal component of the applied force minus the friction force: $F_{net} = F \cos \theta - f_{k} = F \cos \theta - \mu_{K}(mg - F \sin \theta)$.
$5$. Using Newton's second law,$F_{net} = ma$,we get $ma = F \cos \theta - \mu_{K}(mg - F \sin \theta)$.
$6$. Dividing by mass $m$,the acceleration is $a = \frac{F}{m} \cos \theta - \mu_{K}(g - \frac{F}{m} \sin \theta)$.

(B) The frictional force acting on the block is $f = \mu N = \mu mg$.
Using Newton's second law,the retardation $a$ is given by $ma = -\mu mg$,so $a = -\mu g$.
Given $\mu = 0.5$ and $g = 9.8\, m/s^2$,the acceleration is $a = -0.5 \times 9.8 = -4.9\, m/s^2$.
Using the kinematic equation $v^2 = u^2 + 2as$,where final velocity $v = 0$ and initial velocity $u = 9.8\, m/s$:
$0 = (9.8)^2 + 2(-4.9)s$
$9.8s = 9.8 \times 9.8$
$s = 9.8\, m$.

(D) Let $M = 40 \,kg$ be the mass of the block on the surface and $m = 4 \,kg$ be the suspended mass. Let $a$ be the acceleration of the system and $T$ be the tension in the string.
For the suspended mass $m$,the equation of motion is:
$mg - T = ma$
$4(10) - T = 4a \implies 40 - T = 4a$ --- $(1)$
For the block $M$ on the surface,the kinetic friction force is $f_k = \mu_k N = \mu_k Mg = 0.02 \times 40 \times 10 = 8 \,N$.
The equation of motion for the block is:
$T - f_k = Ma$
$T - 8 = 40a$ --- $(2)$
Adding equations $(1)$ and $(2)$:
$(40 - T) + (T - 8) = 4a + 40a$
$32 = 44a$
$a = \frac{32}{44} = \frac{8}{11} \,m/s^2$.

(B) When the bag is dropped on the belt,it experiences a kinetic frictional force $f_k = \mu N = \mu mg$.
This force provides an acceleration $a = f_k / m = \mu g$ to the bag in the direction of the belt's motion.
Given $\mu = 0.4$ and $g = 10\,m/s^2$,the acceleration is $a = 0.4 \times 10 = 4\,m/s^2$.
In the frame of the belt,the initial velocity of the bag is $u = 2\,m/s$ (relative to the belt) and the final velocity is $v = 0$ (when it stops slipping).
Using the equation of motion $v^2 = u^2 - 2as$ (where $s$ is the distance relative to the belt):
$0^2 = 2^2 - 2(4)s$
$0 = 4 - 8s$
$8s = 4$
$s = 0.5\,m$.

(B) The correct answer is $B$.
When Meena applies the brakes,the brake pads press against the wheel rim,preventing the wheel from rotating. However,the bicycle continues to move forward due to inertia. This causes the tyre to slide against the road surface. The road exerts a frictional force on the tyre in the direction opposite to the motion of the bicycle. This external frictional force is responsible for slowing down the bicycle.

(C) Let the mass per unit length of the rope be $\lambda$.
The length of the rope hanging is $f l$,and the length of the rope on the table is $(1-f)l$.
The weight of the hanging part of the rope,which acts as the pulling force $F$,is:
$F = (f l) \lambda g$
The normal reaction $N$ exerted by the table on the part of the rope lying on the table is:
$N = ((1-f)l) \lambda g$
The maximum static friction force $f_s$ acting on the rope is:
$f_s = \mu N = \mu (1-f) l \lambda g$
For the rope to be at rest (in equilibrium),the pulling force must be equal to the limiting friction:
$F = f_s$
$f l \lambda g = \mu (1-f) l \lambda g$
Dividing both sides by $l \lambda g$:
$f = \mu (1-f)$
$f = \mu - \mu f$
$f + \mu f = \mu$
$f(1+\mu) = \mu$
$f = \frac{\mu}{1+\mu}$
To match the given options,we can rewrite this as:
$f = \frac{1}{\frac{1+\mu}{\mu}} = \frac{1}{\frac{1}{\mu} + 1} = \frac{1}{1 + \frac{1}{\mu}}$
Thus,the correct option is $(c)$.

(A) For the block,the final velocity $v = 0$,time $t = 2 \, s$,and displacement $s = 1 \, m$.
Using the equation of motion $v = u + at$,we get:
$0 = u + a \times 2 \Rightarrow a = -u/2$.
Using the equation $v^2 - u^2 = 2as$,we get:
$0 - u^2 = 2 \times (-u/2) \times 1 \Rightarrow -u^2 = -u \Rightarrow u(u - 1) = 0$.
Since $u \neq 0$,the initial velocity is $u = 1 \, m/s$.
The acceleration is $a = -u/2 = -0.5 \, m/s^2$.
The force of kinetic friction is $f_k = \mu_k mg = m|a|$.
Thus,$\mu_k g = |a| \Rightarrow \mu_k = 0.5 / 10 = 0.05$.
In this calculation,we have ignored air resistance and other dissipative forces. In a real-world scenario,these additional forces would contribute to the deceleration,meaning the friction force required to stop the box in $2 \, s$ would be less than the value calculated solely based on kinetic friction. Therefore,$\mu_k$ must be less than $0.05$.

(A) The block is already in motion with a velocity $v = 4 \ m/s$ to the left.
Since the block is moving,the friction acting on it is kinetic friction.
The formula for kinetic friction is $f_k = \mu_k N$,where $N$ is the normal force.
For a block on a horizontal surface,$N = mg$.
Given $m = 10 \ kg$,$g = 10 \ m/s^2$,and $\mu_k = 0.6$:
$f_k = 0.6 \times 10 \times 10 = 60 \ N$.
The direction of kinetic friction is always opposite to the direction of motion.
Since the block is moving to the left,the kinetic frictional force acts to the right with a magnitude of $60 \ N$.

(B) The system of blocks $A$ and $B$ is moving with a constant velocity of $10 \, m/s$.
Since the velocity is constant,the acceleration of the system is $a = 0 \, m/s^2$.
According to Newton's second law of motion,the net external force acting on the system in the horizontal direction must be zero.
Let $F_{ext} = 20 \, N$ be the applied force on block $B$ in the rightward direction.
Let $f_g$ be the friction force applied by the ground on block $B$ acting in the leftward direction.
For the system to be in equilibrium (constant velocity),the net force in the horizontal direction must be zero:
$F_{ext} - f_g = 0$
$20 \, N - f_g = 0$
$f_g = 20 \, N$.
Therefore,the friction force applied by the ground on block $B$ is $20 \, N$.

(A) Given:
Mass of the block,$m = 10 \, kg$
Acceleration,$a = 2 \, m/s^2$
Applied force,$F = 40 \, N$
Acceleration due to gravity,$g = 10 \, m/s^2$ (taking $g = 10 \, m/s^2$ for standard calculation)
The normal force $N$ acting on the block is $N = mg = 10 \times 10 = 100 \, N$.
The kinetic friction force is given by $f_k = \mu_k N = \mu_k \times 100$.
According to Newton's second law of motion,the net force acting on the block is:
$F - f_k = ma$
$40 - 100 \mu_k = 10 \times 2$
$40 - 100 \mu_k = 20$
$100 \mu_k = 40 - 20$
$100 \mu_k = 20$
$\mu_k = \frac{20}{100} = 0.2$
Thus,the coefficient of kinetic friction is $0.2$.

(D) When the brakes are applied,the wheels of the cycle stop rotating and start sliding over the road surface.
In this state,the friction acting between the tires and the road is sliding friction.
Under normal conditions,the cycle moves by rolling,where rolling friction acts.
Since sliding friction is significantly greater than rolling friction,it becomes difficult to move the cycle when the brakes are applied.
Therefore,the correct option is $D$.