Kinetic Friction and Motion on Rough Horizontal Surface Questions in English

(D) The child is sliding down the rope,so the gravitational force acts downwards and the frictional force acts upwards.
Given:
Mass of the child,$m = 25 \,kg$
Frictional force,$f_s = 200 \,N$
Acceleration due to gravity,$g = 10 \,m/s^2$
According to Newton's second law of motion:
$mg - f_s = ma$
Substituting the values:
$(25 \times 10) - 200 = 25 \times a$
$250 - 200 = 25a$
$50 = 25a$
$a = 2 \,m/s^2$
Thus,the acceleration of the child is $2 \,m/s^2$.

(B) Given: Mass $m = 1 \,kg$,initial velocity $u = 8 \,m/s$,final velocity $v = 0 \,m/s$,time $t = 10 \,s$.
First,we calculate the retardation (deceleration) $a$ using the equation of motion: $v = u + at$.
$0 = 8 + a(10) \implies a = -0.8 \,m/s^2$.
The frictional force $f$ acting on the object is $f = ma = 1 \times 0.8 = 0.8 \,N$.
To keep the object moving at a constant velocity of $8 \,m/s$,the applied force $F$ must exactly balance the frictional force $f$.
Therefore,$F = f = 0.8 \,N$.

(B) Given:
Initial velocity $u = 4 \, m/s$
Final velocity $v = 0 \, m/s$
Time $t = 8 \, s$
Frictional force $f = 10 \, N$
First,calculate the acceleration (deceleration) using the equation of motion $v = u + at$:
$0 = 4 + a(8)$
$8a = -4$
$a = -0.5 \, m/s^2$
The magnitude of acceleration is $0.5 \, m/s^2$.
Using Newton's Second Law,$f = ma$:
$10 = m \times 0.5$
$m = \frac{10}{0.5}$
$m = 20 \, kg$
Therefore,the mass of the box is $20 \, kg$.

(D) The correct answer is $D$.
When the engine of the car rotates the wheels,the wheels exert a backward force on the road.
According to Newton's $3^{rd}$ law of motion,the road exerts an equal and opposite forward force on the wheels.
This forward frictional force provided by the road is the external force that causes the car to accelerate on the horizontal surface.

(A) The retardation $a$ of the block is given by $a = \frac{dv}{dt} = -\mu g$.
Integrating this expression,we get $\Delta v = v_f - v_i = -g \int_{0}^{4} \mu \,dt$.
The term $\int_{0}^{4} \mu \,dt$ represents the area under the $\mu-t$ curve from $t = 0$ to $t = 4 \,s$.
The area consists of a rectangle from $t = 0$ to $2 \,s$ and a trapezoid from $t = 2$ to $4 \,s$.
Area $= (2 \times 0.5) + \frac{1}{2} \times (0.5 + 0.3) \times (4 - 2) = 1 + 0.8 = 1.8$.
Now,substituting the values into the velocity equation:
$v_f - 20 = -10 \times (1.8) = -18$.
$v_f = 20 - 18 = 2 \,m/s$.

(B) Given: Mass $m = 40\,kg$,coefficient of static friction $\mu_{s} = 0.5$,coefficient of kinetic friction $\mu_{k} = 0.4$,and acceleration due to gravity $g = 10\,m/s^2$.
The force $P$ is just enough to start the motion,so it must be equal to the limiting friction:
$P = f_{s,max} = \mu_{s} N = \mu_{s} mg$.
Once the body starts moving,the kinetic friction $f_{k}$ acts on it:
$f_{k} = \mu_{k} N = \mu_{k} mg$.
The net force $F_{net}$ acting on the body is:
$F_{net} = P - f_{k} = \mu_{s} mg - \mu_{k} mg = m(\mu_{s} - \mu_{k})g$.
Using Newton's second law,$F_{net} = ma$:
$ma = m(\mu_{s} - \mu_{k})g$.
Therefore,the acceleration $a$ is:
$a = (\mu_{s} - \mu_{k})g$.
Substituting the values:
$a = (0.5 - 0.4) \times 10 = 0.1 \times 10 = 1\,m/s^2$.

(D) Given: Mass $m = 10\,kg$,initial velocity $u = 20\,m/s$,final velocity $v = 0\,m/s$,time $t = 5\,s$,and $g = 10\,m/s^2$.
Using the first equation of motion,$v = u + at$,we find the acceleration $a$:
$0 = 20 + a(5) \implies 5a = -20 \implies a = -4\,m/s^2$.
The retarding force is provided by friction,so $F_f = ma = -\mu mg$.
Thus,$-\mu g = a$.
$-\mu(10) = -4$.
$\mu = \frac{4}{10} = 0.4$.

(C) Using the equation of motion $S = ut + \frac{1}{2}at^2$,where $S = 50\,m$,$u = 0\,m/s$,and $t = 10\,s$:
$50 = 0 + \frac{1}{2} \times a \times (10)^2$
$50 = 50a$
$a = 1\,m/s^2$
Now,applying Newton's second law,the net force is $F - f_k = ma$,where $f_k = \mu_k N = \mu_k mg$:
$30 - \mu_k \times 5 \times 10 = 5 \times 1$
$30 - 50\mu_k = 5$
$50\mu_k = 25$
$\mu_k = \frac{25}{50} = 0.50$

(C) Given:
Mass $m = 100 \ kg$
Distance $s = 10 \ m$
Coefficient of friction $\mu = 0.4$
Acceleration due to gravity $g = 10 \ m/s^2$ (assuming standard value).
The frictional force $f$ is given by the formula:
$f = \mu N = \mu mg$
$f = 0.4 \times 100 \times 10 = 400 \ N$
The work done against friction $W$ is given by:
$W = f \times s$
$W = 400 \times 10 = 4000 \ J$
Therefore,the work done against friction is $4000 \ J$.

(C) The mass of the trolley is $m_1 = 20 \text{ kg}$ and the mass of the hanging block is $m_2 = 6 \text{ kg}$.
The normal force on the trolley is $N = m_1 g = 20 \times 10 = 200 \text{ N}$.
The kinetic friction force is $f_k = \mu_k N = 0.04 \times 200 = 8 \text{ N}$.
The driving force is the weight of the hanging block,$F = m_2 g = 6 \times 10 = 60 \text{ N}$.
Applying Newton's second law to the system: $F - f_k = (m_1 + m_2) a$.
$60 - 8 = (20 + 6) a$.
$52 = 26 a$.
$a = \frac{52}{26} = 2 \text{ m/s}^2$.

(B) The force of kinetic friction $f_k$ 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.
For a box on a horizontal surface, the normal force $N$ is equal to the weight of the box, $mg$.
Given:
Mass $m = 50 \,kg$
Coefficient of kinetic friction $\mu_k = 0.3$
Acceleration due to gravity $g = 9.8 \,m/s^2$
Calculating the normal force:
$N = mg = 50 \times 9.8 = 490 \,N$
Calculating the kinetic friction force:
$f_k = 0.3 \times 490 = 147 \,N$

(B) For the system to be in equilibrium (motion stopped),the forces must be balanced.
For mass $m_1$ hanging vertically: $T = m_1 g = 10 \times 10 = 100 \text{ N}$.
For the combined mass $(m + m_2)$ on the horizontal surface,the limiting friction $f_L$ must balance the tension $T$:
$f_L = T$
$\mu N = T$
$\mu (m + m_2) g = 100$
$0.15 \times (m + 20) \times 10 = 100$
$1.5 (m + 20) = 100$
$m + 20 = \frac{100}{1.5} = \frac{1000}{15} = \frac{200}{3}$
$m = \frac{200}{3} - 20 = \frac{200 - 60}{3} = \frac{140}{3} \text{ kg}$.

(D) The forces acting on the fireman are his weight $W = mg$ acting downwards and the frictional force $f = \mu R$ acting upwards,where $R$ is the normal force exerted by the fireman on the pole.
Given:
Mass $m = 60 \ kg$
Normal force $R = 600 \ N$
Coefficient of friction $\mu = 0.5$
Acceleration due to gravity $g = 10 \ m/s^2$
The net force $F_{\text{net}}$ acting downwards is given by:
$F_{\text{net}} = mg - f = mg - \mu R$
Using Newton's second law,$F_{\text{net}} = ma$:
$ma = mg - \mu R$
$a = \frac{mg - \mu R}{m}$
Substituting the values:
$a = \frac{60 \times 10 - 0.5 \times 600}{60}$
$a = \frac{600 - 300}{60}$
$a = \frac{300}{60} = 5 \ m/s^2$
Thus,the fireman slides down with an acceleration of $5 \ m/s^2$.

(A) Let $F$ be the maximum value of the force applied such that the block of mass $m = \sqrt{3} \ kg$ does not move on the rough surface.
$R$ is the normal reaction force exerted by the surface on the block.
Resolving the forces vertically:
$R = F \sin 60^{\circ} + mg$
Resolving the forces horizontally,the limiting friction $f = \mu R$ must balance the horizontal component of the applied force $F \cos 60^{\circ}$:
$f = F \cos 60^{\circ}$
$\mu R = F \cos 60^{\circ}$
Substituting the expression for $R$:
$\mu (F \sin 60^{\circ} + mg) = F \cos 60^{\circ}$
$\mu F \sin 60^{\circ} + \mu mg = F \cos 60^{\circ}$
$F (\cos 60^{\circ} - \mu \sin 60^{\circ}) = \mu mg$
$F = \frac{\mu mg}{\cos 60^{\circ} - \mu \sin 60^{\circ}}$
Given $\mu = \frac{1}{2 \sqrt{3}}$,$m = \sqrt{3} \ kg$,and $g = 10 \ m/s^2$:
$F = \frac{(\frac{1}{2 \sqrt{3}}) \times \sqrt{3} \times 10}{\cos 60^{\circ} - (\frac{1}{2 \sqrt{3}}) \sin 60^{\circ}}$
$F = \frac{5}{\frac{1}{2} - (\frac{1}{2 \sqrt{3}} \times \frac{\sqrt{3}}{2})} = \frac{5}{\frac{1}{2} - \frac{1}{4}} = \frac{5}{\frac{1}{4}} = 20 \ N$
Therefore,the maximum value of the force is $20 \ N$.

(A) The surface exerts two forces on the block: the normal force $(N)$ acting vertically upwards and the kinetic frictional force $(f_{k})$ acting horizontally opposite to the direction of motion.
$1$. The normal force is $N = mg$.
$2$. The kinetic frictional force is $f_{k} = \mu_{k} N = \mu_{k} mg$.
$3$. The net force $(F_{net})$ exerted by the surface is the vector sum of the normal force and the frictional force:
$F_{net} = \sqrt{N^{2} + f_{k}^{2}}$
$F_{net} = \sqrt{(mg)^{2} + (\mu_{k} mg)^{2}}$
$F_{net} = \sqrt{(mg)^{2} (1 + \mu_{k}^{2})}$
$F_{net} = mg \sqrt{1 + \mu_{k}^{2}}$
$F_{net} = mg(1 + \mu_{k}^{2})^{1/2}$
Therefore,option $A$ is correct.

(C) The initial momentum of the block is $P = Mu$,where $u$ is the initial velocity. Thus,$u = \frac{P}{M}$.
Since the surface is rough,the frictional force $f = \mu N = \mu Mg$ acts on the block.
The retardation (deceleration) $a$ is given by $a = \frac{f}{M} = \frac{\mu Mg}{M} = \mu g$.
Using the kinematic equation $v^{2} = u^{2} - 2as$,where $v = 0$ (final velocity when it stops):
$0 = u^{2} - 2as$
$2as = u^{2}$
$s = \frac{u^{2}}{2a}$
Substituting $u = \frac{P}{M}$ and $a = \mu g$:
$s = \frac{(\frac{P}{M})^{2}}{2 \mu g} = \frac{P^{2}}{M^{2} \cdot 2 \mu g} = \frac{P^{2}}{2 \mu M^{2} g}$.

(C) Given: Initial velocity $u = 4 \,m/s$, final velocity $v = 0 \,m/s$, coefficient of kinetic friction $\mu = 0.2$, and acceleration due to gravity $g = 10 \,m/s^2$.
The frictional force $f$ acting on the body is $f = \mu N = \mu mg$.
According to Newton's second law, the retardation $a$ is given by $a = -f/m = -\mu g$.
Substituting the values: $a = -(0.2) \times 10 = -2 \,m/s^2$.
Using the kinematic equation $v^2 = u^2 + 2as$:
$0^2 = (4)^2 + 2(-2)s$
$0 = 16 - 4s$
$4s = 16$
$s = 4 \,m$.
Therefore, the distance travelled by the body before coming to rest is $4 \,m$.

(A) The initial momentum of the vehicle is $P = Mv$,so the initial velocity is $u = \frac{P}{M}$.
The final velocity $v = 0$ as the vehicle comes to a stop.
The frictional force acting on the vehicle is $f = \mu N = \mu Mg$,where $N = Mg$ is the normal reaction.
Using Newton's second law,the retardation $a$ is given by $a = -\frac{f}{M} = -\frac{\mu Mg}{M} = -\mu g$.
Using the kinematic equation $v^{2} - u^{2} = 2as$,where $s$ is the stopping distance:
$0^{2} - (\frac{P}{M})^{2} = 2(-\mu g)s$
$-\frac{P^{2}}{M^{2}} = -2\mu gs$
$s = \frac{P^{2}}{2\mu g M^{2}}$.

(D) From the velocity-time graph,the acceleration $a$ of the block is the slope of the line:
$a = \frac{v - u}{t} = \frac{20 - 0}{3} = \frac{20}{3} \text{ m/s}^2$
Since the block is moving,the friction acting on it is kinetic friction,$f_k = \mu_k N = \mu_k mg$.
Given $\mu_k = 0.25$ and $g = 10 \text{ m/s}^2$,the kinetic friction is $f_k = 0.25 \times m \times 10 = 2.5m$.
Applying Newton's second law of motion,$F - f_k = ma$:
$20 - 2.5m = m \times \frac{20}{3}$
$20 = m \left( \frac{20}{3} + 2.5 \right)$
$20 = m \left( \frac{20 + 7.5}{3} \right) = m \left( \frac{27.5}{3} \right)$
$m = \frac{20 \times 3}{27.5} = \frac{60}{27.5} \approx 2.18 \text{ kg}$.
Rounding to the nearest value,the mass is $2.2 \text{ kg}$.

$(A)$ Given: Mass of the body,$m = 10 \,kg$. Coefficient of kinetic friction,$\mu_k = 0.5$. Applied horizontal force,$F = 60 \,N$. Acceleration due to gravity,$g = 10 \,m/s^2$.
The normal force $N$ acting on the body is $N = mg = 10 \times 10 = 100 \,N$.
The kinetic frictional force $f_k$ is given by $f_k = \mu_k N = 0.5 \times 100 = 50 \,N$.
According to Newton's second law,the net force $F_{net}$ acting on the body is $F_{net} = F - f_k = 60 \,N - 50 \,N = 10 \,N$.
The acceleration $a$ of the body is $a = F_{net} / m = 10 \,N / 10 \,kg = 1 \,m/s^2$.

(C) The body is placed on a moving belt. Initially, the body is at rest relative to the ground, but it has a velocity relative to the belt. The kinetic friction force $f_k$ acts on the body to oppose this relative motion.
$f_k = \mu_k N = \mu_k mg$
Using Newton's second law, the deceleration $a$ of the body relative to the belt is:
$ma = \mu_k mg$
$a = \mu_k g = 0.2 \times 10 = 2 \,m \,s^{-2}$
The initial velocity of the body relative to the belt is $u = 2 \,m \,s^{-1}$. The body comes to rest relative to the belt when its final velocity $v = 0$.
Using the equation of motion $v^2 = u^2 - 2as$:
$0^2 = (2)^2 - 2(2)s$
$4 = 4s$
$s = 1 \,m$
Thus, the distance travelled by the body before coming to rest relative to the belt is $1 \,m$.

(B) Given: Mass $m = 2 \ kg$,Applied force $F = 20 \ N$,Acceleration $a = 7 \ m \ s^{-2}$,Acceleration due to gravity $g = 10 \ m \ s^{-2}$.
According to Newton's second law of motion,the net force acting on the block is $F_{net} = F - f_k = ma$,where $f_k$ is the kinetic frictional force.
The kinetic frictional force is given by $f_k = \mu_k N$,where $N$ is the normal reaction force. On a horizontal surface,$N = mg = 2 \ kg \times 10 \ m \ s^{-2} = 20 \ N$.
Substituting the values into the equation $F - f_k = ma$:
$20 - f_k = 2 \times 7$
$20 - f_k = 14$
$f_k = 20 - 14 = 6 \ N$.
Now,using $f_k = \mu_k N$:
$6 = \mu_k \times 20$
$\mu_k = 6 / 20 = 0.3$.
Therefore,the coefficient of kinetic friction is $0.3$.

(B) Given: Initial velocity $u = 10 \ ms^{-1}$,Final velocity $v = 0 \ ms^{-1}$,Distance $S = 12.5 \ m$,Acceleration due to gravity $g = 10 \ ms^{-2}$.
The frictional force $f$ acting on the block is $f = \mu N = \mu mg$,where $\mu$ is the coefficient of kinetic friction.
The retardation $a$ produced by this frictional force is $a = \frac{f}{m} = \frac{\mu mg}{m} = \mu g$.
Using the equation of motion $v^2 = u^2 - 2aS$,where $v = 0$:
$0 = u^2 - 2aS$
$u^2 = 2aS$
$S = \frac{u^2}{2a} = \frac{u^2}{2 \mu g}$.
Substituting the given values:
$12.5 = \frac{(10)^2}{2 \times \mu \times 10}$
$12.5 = \frac{100}{20 \mu}$
$12.5 = \frac{5}{\mu}$
$\mu = \frac{5}{12.5} = \frac{50}{125} = 0.4$.
Therefore,the coefficient of kinetic friction is $0.4$.

(A) Given: Mass $m = 5 \ kg$,coefficient of friction $\mu = 0.5$,applied force $F = 60 \ N$,and $g = 10 \ ms^{-2}$.
First,calculate the limiting frictional force $f_l$ acting on the block:
$f_l = \mu N = \mu mg = 0.5 \times 5 \times 10 = 25 \ N$.
Since the applied force $F = 60 \ N$ is greater than the limiting frictional force $f_l = 25 \ N$,the block will move.
The net force acting on the block is $F_{net} = F - f_l = 60 - 25 = 35 \ N$.
Using Newton's second law,$F_{net} = ma$,the acceleration $a$ is:
$a = \frac{F_{net}}{m} = \frac{35}{5} = 7 \ ms^{-2}$.

(B) The frictional force acting on the particle is $f_r = \mu m g$.
Since the particle is moving,the acceleration is $a = \frac{-f_r}{m} = \frac{-\mu m g}{m} = -\mu g$.
The time taken to stop is given by $v = u + at$. Setting $v = 0$,we get $0 = u - \mu g t$,so $t = \frac{u}{\mu g}$.
The work done by friction is equal to the change in kinetic energy: $W_f = \Delta K = 0 - \frac{1}{2} m u^2 = -\frac{1}{2} m u^2$.
The average power imparted by friction is $P_{av} = \frac{|W_f|}{t} = \frac{\frac{1}{2} m u^2}{\frac{u}{\mu g}} = \frac{1}{2} \mu m g u$.

(D) Given: Mass $m = 5 \,kg$, initial velocity $u = 4 \,ms^{-1}$, final velocity $v = 0 \,ms^{-1}$, time $t = 2 \,s$, and $g = 10 \,ms^{-2}$.
First, calculate the retardation $a$ using the equation of motion $v = u + at$:
$0 = 4 + a(2) \Rightarrow 2a = -4 \Rightarrow a = -2 \,ms^{-2}$.
The magnitude of retardation is $|a| = 2 \,ms^{-2}$.
The frictional force $f$ provides this retardation, so $f = ma$.
Also, the frictional force is given by $f = \mu N = \mu mg$.
Equating the two expressions for $f$:
$ma = \mu mg \Rightarrow \mu = \frac{a}{g}$.
Substituting the values:
$\mu = \frac{2}{10} = 0.2$.

(A) Given: Mass $m = 1 \,kg$, initial velocity $u = 10 \,ms^{-1}$, final velocity $v = 0 \,ms^{-1}$, coefficient of kinetic friction $\mu = 0.4$, and $g = 10 \,ms^{-2}$.
When the force is removed, the only horizontal force acting on the body is the kinetic frictional force $f_k = \mu N = \mu mg$.
According to Newton's second law, the retardation $a$ is given by $a = \frac{f_k}{m} = \frac{\mu mg}{m} = \mu g$.
Substituting the values: $a = 0.4 \times 10 = 4 \,ms^{-2}$.
Using the first equation of motion $v = u - at$ (where $a$ is retardation):
$0 = 10 - 4t$
$4t = 10$
$t = \frac{10}{4} = 2.5 \,s$.
Thus, the body comes to rest in $2.5 \,s$.

(B) Given: Mass $m = 10 \,kg$, Coefficient of friction $\mu = 0.3$, Applied force $F = 50 \,N$, Acceleration due to gravity $g = 10 \,ms^{-2}$.
First, calculate the limiting frictional force $f_l = \mu mg = 0.3 \times 10 \times 10 = 30 \,N$.
Since the applied force $F = 50 \,N$ is greater than the limiting frictional force $f_l = 30 \,N$, the body will move.
The net force acting on the body is $F_{net} = F - f_l = 50 \,N - 30 \,N = 20 \,N$.
Using Newton's second law, $F_{net} = ma$, we get $20 = 10 \times a$.
Therefore, the acceleration $a = 2 \,ms^{-2}$.

(B) Initial speed of the body on the horizontal rough surface,$u = 10 \,ms^{-1}$.
Final velocity,$v = 4 \,ms^{-1}$ and time,$t = 2 \,s$.
If $a$ is the retardation due to friction,then by using the equation of motion:
$v = u - at$
$4 = 10 - a \times 2$
$2a = 6 \Rightarrow a = 3 \,ms^{-2}$.
According to Newton's second law,the frictional force is given by $f_k = ma$.
Since $f_k = \mu_k N = \mu_k mg$,we have:
$\mu_k mg = ma$
$\mu_k = \frac{a}{g} = \frac{3}{10} = 0.3$.

(A) Given: Mass of the cylinder, $m = 12 \,kg$. Initial velocity, $u = 20 \,m/s$. Coefficient of kinetic friction, $\mu = 0.5$. Acceleration due to gravity, $g = 10 \,m/s^2$.
The frictional force acting on the cylinder is $f = \mu N = \mu mg$.
According to Newton's second law, $f = ma$, so $ma = \mu mg$, which gives the retardation $a = -\mu g$.
Substituting the values, $a = -0.5 \times 10 = -5 \,m/s^2$.
Using the third equation of motion, $v^2 = u^2 + 2as$, where $v = 0$ (final velocity at rest):
$0 = (20)^2 + 2(-5)s$
$0 = 400 - 10s$
$10s = 400$
$s = 40 \,m$.
Therefore, the cylinder travels a distance of $40 \,m$ before stopping.

(A) Given: Weight of the body,$W = 64 \ N$.
Coefficient of static friction,$\mu_s = 0.8$.
Coefficient of dynamic (kinetic) friction,$\mu_d = 0.6$.
The force $F$ required to just start the motion is equal to the limiting friction: $F = \mu_s \times W = 0.8 \times 64 \ N$.
Once the body starts moving,the kinetic friction acting on it is $f_k = \mu_d \times W = 0.6 \times 64 \ N$.
The net force acting on the body is $F_{net} = F - f_k = (0.8 \times 64) - (0.6 \times 64) = 64(0.8 - 0.6) = 64 \times 0.2 \ N$.
Using Newton's second law,$F_{net} = m \times a$,where mass $m = \frac{W}{g} = \frac{64}{g}$.
Substituting the values: $\frac{64}{g} \times a = 64 \times 0.2$.
Therefore,$a = 0.2 \ g$.

(B) The weight of the block acts downwards: $w = mg = 2 \times 10 = 20 \,N$.
The applied force $F = 90 \,N$ is at an angle of $37^{\circ}$ to the horizontal.
The vertical component of the force is $F_V = F \sin 37^{\circ} = 90 \times \frac{3}{5} = 54 \,N$ (upwards).
The horizontal component of the force is $F_H = F \cos 37^{\circ} = 90 \times \frac{4}{5} = 72 \,N$ (towards the wall).
This horizontal component acts as the normal force $N$ exerted by the wall on the block: $N = 72 \,N$.
The maximum frictional force is $f_k = \mu N = 0.25 \times 72 = 18 \,N$.
Since the block is moving upwards, the friction acts downwards.
The net force in the vertical direction is $F_{\text{net}} = F_V - w - f_k = 54 - 20 - 18 = 16 \,N$.
Using Newton's second law, $F_{\text{net}} = ma$, we get $16 = 2 \times a$.
Therefore, $a = 8 \,ms^{-2}$.

(B) Let $\lambda$ be the linear mass density of the chain. The total length of the chain is $L$. Let $l^{\prime}$ be the length of the chain hanging over the edge.
Then,the length of the chain remaining on the table is $(L - l^{\prime})$.
The mass of the hanging part is $m_h = \lambda l^{\prime}$,and the mass of the part on the table is $m_t = \lambda (L - l^{\prime})$.
The force pulling the chain down is the weight of the hanging part: $F_g = m_h g = \lambda l^{\prime} g$.
The maximum static frictional force acting on the part on the table is $f_{max} = \mu N = \mu m_t g = \mu \lambda (L - l^{\prime}) g$.
For the chain to be in equilibrium,the pulling force must be equal to the maximum frictional force:
$\lambda l^{\prime} g = \mu \lambda (L - l^{\prime}) g$
$l^{\prime} = \mu (L - l^{\prime})$
$l^{\prime} = \mu L - \mu l^{\prime}$
$l^{\prime} (1 + \mu) = \mu L$
$l^{\prime} = \frac{\mu L}{(1 + \mu)}$

(D) Given: Weight $W = 64 \,N$, coefficient of static friction $\mu_s = 0.6$, coefficient of kinetic (dynamic) friction $\mu_k = 0.4$.
The force applied to start the motion is equal to the limiting friction: $F = f_{s,max} = \mu_s N = \mu_s W$.
Since $W = mg$, we have $F = 0.6 \times mg$.
Once the body starts moving, the kinetic friction acting on it is $f_k = \mu_k N = \mu_k mg = 0.4 \times mg$.
The net force acting on the body is $F_{net} = F - f_k = 0.6 mg - 0.4 mg = 0.2 mg$.
According to Newton's second law, $F_{net} = ma$.
Therefore, $ma = 0.2 mg$, which gives $a = 0.2 g$.

(B) The force of kinetic friction acting on a block is given by $f_k = \mu_k N = \mu_k mg$.
For block $A$: Mass $m_A = 2 \ kg$,Force $F = 20 \ N$. Friction $f_A = 0.3 \times 2 \times 10 = 6 \ N$. Net force $F_{net,A} = F - f_A = 20 - 6 = 14 \ N$. Acceleration $a_A = F_{net,A} / m_A = 14 / 2 = 7 \ m \ s^{-2}$.
For block $B$: Mass $m_B = 4 \ kg$,Force $F = 20 \ N$. Friction $f_B = 0.3 \times 4 \times 10 = 12 \ N$. Net force $F_{net,B} = F - f_B = 20 - 12 = 8 \ N$. Acceleration $a_B = F_{net,B} / m_B = 8 / 4 = 2 \ m \ s^{-2}$.
The ratio of accelerations is $a_A : a_B = 7 : 2$.

(D) Given: Mass of the boat, $m = 700 \,kg$. Initial speed, $v_1 = 24 \,ms^{-1}$. Final speed, $v_2 = 6 \,ms^{-1}$. Frictional force, $f = 35v$.
According to Newton's second law, the retarding force is $f = -m \frac{dv}{dt}$.
Substituting the given values: $35v = -700 \frac{dv}{dt}$.
Rearranging the terms: $\frac{dv}{v} = -\frac{35}{700} dt = -\frac{1}{20} dt$.
Integrating both sides from $v_1$ to $v_2$ and $0$ to $t$: $\int_{24}^{6} \frac{dv}{v} = -\int_{0}^{t} \frac{1}{20} dt$.
$\ln(\frac{6}{24}) = -\frac{t}{20}$.
$\ln(\frac{1}{4}) = -\frac{t}{20}$.
$-\ln(4) = -\frac{t}{20} \Rightarrow t = 20 \ln(4) = 20 \ln(2^2) = 40 \ln(2)$.
Using $\ln(2) \approx 0.693$, $t = 40 \times 0.693 = 27.72 \,s$.
Rounding to the nearest integer, $t \approx 28 \,s$.

(A) To maintain a constant speed, the engine must exert a force equal to the frictional force acting on the train.
The frictional force $F_f$ is given by $F_f = \mu N$, where $\mu$ is the coefficient of kinetic friction and $N$ is the normal force.
Since the train is on a horizontal track, $N = mg$.
Given: $m = 3 \times 10^6 \,kg$, $\mu = 0.05$, $g = 10 \,m \,s^{-2}$, and velocity $v = 50 \,m \,s^{-1}$.
$F_f = 0.05 \times (3 \times 10^6 \,kg) \times (10 \,m \,s^{-2}) = 0.05 \times 3 \times 10^7 \,N = 1.5 \times 10^6 \,N$.
The power $P$ required is given by $P = F_f \times v$.
$P = (1.5 \times 10^6 \,N) \times (50 \,m \,s^{-1}) = 75 \times 10^6 \,W = 75 \,MW$.

(B) Given,tension in the string $(T) = 27 \,N$.
Mass of the hanging block $(m) = 3 \,kg$.
Let the acceleration of the system be $a$.
For the hanging block of mass $3 \,kg$,the equation of motion is:
$m g - T = m a$
$3 \times 10 - 27 = 3 a$
$30 - 27 = 3 a$
$3 = 3 a$
$a = 1 \,m/s^2$.
For the body of mass $M = 20 \,kg$ moving on the horizontal surface,the equation of motion is:
$T - f_k = M a$
where $f_k = \mu M g$ is the kinetic friction force.
$27 - \mu \times 20 \times 10 = 20 \times 1$
$27 - 200 \mu = 20$
$200 \mu = 27 - 20$
$200 \mu = 7$
$\mu = \frac{7}{200} = 0.035$.
Therefore,the coefficient of kinetic friction is $0.035$.

(C) Let the mass of the man be $m$. The weight of the man is $w = mg$. The forces acting on the man are his weight $mg$ acting downwards and the frictional force $F$ acting upwards.
Since the man is sliding down with an acceleration $a = g/4$,according to Newton's second law of motion:
$mg - F = ma$
Substituting $a = g/4$:
$mg - F = m(g/4)$
$mg - F = mg/4$
$F = mg - mg/4$
$F = 3mg/4$
Since $w = mg$,we get:
$F = 3w/4$

(C) The block is on a horizontal surface. The forces acting on the block are:
$1$. Weight $mg$ acting downwards.
$2$. Normal reaction $N$ acting upwards.
$3$. Applied force $F$ at $45^{\circ}$ to the horizontal.
Resolving $F$ into components: $F_x = F \cos 45^{\circ}$ and $F_y = F \sin 45^{\circ}$.
For vertical equilibrium: $N + F \sin 45^{\circ} = mg$.
So, $N = mg - F \sin 45^{\circ} = \sqrt{2} \times 10 - F \times \frac{1}{\sqrt{2}} = 10\sqrt{2} - \frac{F}{\sqrt{2}}$.
The block starts to move when the horizontal component of the force equals the limiting friction: $F \cos 45^{\circ} = \mu N$.
Substituting the values: $F \times \frac{1}{\sqrt{2}} = 0.25 \times (10\sqrt{2} - \frac{F}{\sqrt{2}})$.
Multiply by $\sqrt{2}$: $F = 0.25 \times (10 \times 2 - F) = 0.25 \times (20 - F)$.
$F = 5 - 0.25F$.
$1.25F = 5$.
$F = \frac{5}{1.25} = 4 \,N$.

(C) The net force acting on the block is given by $F_{net} = F - f_k = ma$,where $f_k = \mu_k mg$ is the kinetic friction force.
Thus,the acceleration is $a = \frac{F - \mu_k mg}{m}$.
For the first case: $6 = \frac{20 - \mu_k mg}{m} \implies 6m = 20 - \mu_k mg$ --- $(i)$
For the second case: $11 = \frac{30 - \mu_k mg}{m} \implies 11m = 30 - \mu_k mg$ --- (ii)
Subtracting equation $(i)$ from equation (ii):
$(11m - 6m) = (30 - \mu_k mg) - (20 - \mu_k mg)$
$5m = 10 \implies m = 2 \ kg$.
Substitute $m = 2 \ kg$ into equation $(i)$:
$6(2) = 20 - \mu_k (2)(10)$
$12 = 20 - 20\mu_k$
$20\mu_k = 8$
$\mu_k = \frac{8}{20} = 0.4$.

(A) The forces acting on the block are as follows:
$1$. The normal reaction $N$ exerted by the wall is $N = F \cos 30^{\circ} = F \frac{\sqrt{3}}{2}$.
$2$. The maximum static friction force is $f_{max} = \mu N = \sqrt{3} \times (F \frac{\sqrt{3}}{2}) = \frac{3}{2} F$.
$3$. The vertical forces acting on the block are the weight $mg$ acting downwards and the vertical component of the applied force $F \sin 30^{\circ}$ acting downwards.
$4$. For the block to be in equilibrium and prevented from falling, the upward friction force must balance the total downward force:
$f = mg + F \sin 30^{\circ}$
$\frac{3}{2} F = (3 \times 10) + F \times \frac{1}{2}$
$\frac{3}{2} F - \frac{1}{2} F = 30$
$F = 30 \,N$.

(A) $1$. First,we determine the normal force $F_N$ acting on the block. The forces in the vertical direction are the upward force $F_2$,the normal force $F_N$,and the downward gravitational force $mg$. Since the block is not moving vertically,the net vertical force is zero:
$F_2 + F_N - mg = 0$
$10 \ N + F_N - (2 \ kg)(10 \ m/s^2) = 0$
$10 \ N + F_N - 20 \ N = 0$
$F_N = 10 \ N$
$2$. Next,we calculate the limiting static friction force $f_{s,max}$:
$f_{s,max} = \mu_s F_N = 0.4 \times 10 \ N = 4.0 \ N$
$3$. The applied horizontal force is $F_1 = 6 \ N$. Since the applied horizontal force $F_1$ is greater than the limiting static friction force $f_{s,max}$ $(6 \ N > 4.0 \ N)$,the block will start to move.
$4$. Once the block is in motion,the frictional force acting on it is the kinetic friction force $f_k$:
$f_k = \mu_k F_N = 0.25 \times 10 \ N = 2.5 \ N$
Therefore,the magnitude of the frictional force acting on the block is $2.5 \ N$.

(B) To move the middle steel plate,the applied force $F$ must overcome the kinetic friction forces acting on both its top and bottom surfaces.
$1$. The normal force $R$ acting on the surfaces is equal to the applied weight,$R = 100 \ N$.
$2$. The friction force on the top surface (between steel and steel) is $f_{SS} = \mu_{SS} \times R = 0.57 \times 100 = 57 \ N$.
$3$. The friction force on the bottom surface (between steel and brass) is $f_{SB} = \mu_{SB} \times R = 0.44 \times 100 = 44 \ N$.
$4$. The total force $F$ required to move the plate is the sum of these two friction forces:
$F = f_{SS} + f_{SB} = 57 \ N + 44 \ N = 101 \ N$.

(B) The forces acting on the block are the applied force $F$ at an angle of $30^{\circ}$ with the horizontal, the weight $mg$ acting downwards, the normal reaction $N$ acting upwards, and the kinetic friction $f_k$ acting to the left.
Resolving the force $F$ into horizontal and vertical components: $F_x = F \cos 30^{\circ}$ and $F_y = F \sin 30^{\circ}$.
For vertical equilibrium: $N + F \sin 30^{\circ} = mg \Rightarrow N = mg - F \sin 30^{\circ}$.
Given $m = 5 \,kg$, $g = 9.8 \,m/s^2$, and $\mu = 0.1$, we have $N = (5 \times 9.8) - F \sin 30^{\circ} = 49 - 0.5F$.
The kinetic friction is $f_k = \mu N = 0.1(49 - 0.5F) = 4.9 - 0.05F$.
The net force in the horizontal direction is $F_{\text{net}} = F \cos 30^{\circ} - f_k = ma$.
Substituting the values: $F(\frac{\sqrt{3}}{2}) - (4.9 - 0.05F) = 5 \times 3$.
$0.866F - 4.9 + 0.05F = 15$.
$0.916F = 19.9$.
$F = \frac{19.9}{0.916} \approx 21.73 \,N$.
This value is closest to $22 \,N$.

(C) Given: Applied force $F_{\text{app}} = 5 \ N$,mass $m = 0.5 \ kg$,coefficient of kinetic friction $\mu_k = 0.2$,time $t = 4 \ s$,initial velocity $u = 0$,acceleration due to gravity $g = 10 \ m/s^2$.
The kinetic frictional force is given by $f_k = \mu_k N = \mu_k mg$.
$f_k = 0.2 \times 0.5 \times 10 = 1 \ N$.
The net force acting on the block is $F_{\text{net}} = F_{\text{app}} - f_k = 5 - 1 = 4 \ N$.
The acceleration of the block is $a = \frac{F_{\text{net}}}{m} = \frac{4}{0.5} = 8 \ m/s^2$.
The velocity of the block after $4 \ s$ is $v = u + at = 0 + (8 \times 4) = 32 \ m/s$.
The kinetic energy of the block is $K.E. = \frac{1}{2} mv^2 = \frac{1}{2} \times 0.5 \times (32)^2 = 0.25 \times 1024 = 256 \ J$.

(A) The initial kinetic energy of the vehicle is given by $K = \frac{p^2}{2M}$.
When the engine is switched off,the only force acting to stop the vehicle is the kinetic friction force $f_k = \mu_k M g$.
The work done by the friction force in stopping the vehicle over a distance $s$ is equal to the change in kinetic energy.
$|W| = f_k \cdot s = \mu_k M g s$.
According to the work-energy theorem,the work done by friction equals the initial kinetic energy:
$\mu_k M g s = \frac{p^2}{2M}$.
Solving for $s$,we get:
$s = \frac{p^2}{2 M^2 \mu_k g}$.