Block on Block System, psudo force and Constrained Motion In Friction Questions in English

(A) Let $m_A$ be the mass of body $A$ and $m_B$ be the mass of body $B$. The table is smooth,so there is no friction between $B$ and the table.
When an acceleration $a$ is given to body $B$,a pseudo force $F_p = m_A a$ acts on body $A$ in the backward direction relative to $B$.
The maximum static friction force available between $A$ and $B$ is $f_{max} = \mu N = \mu m_A g$,where $N = m_A g$ is the normal force between $A$ and $B$.
Slipping will occur when the pseudo force exceeds the maximum static friction force.
Therefore,the condition for slipping is $m_A a > \mu m_A g$.
Simplifying this,we get $a > \mu g$.
Thus,the minimum acceleration required for slipping to occur is $a = \mu g$.

(A) $1$. Calculate the limiting friction between the block $(m_A = 10 \,kg)$ and the slab $(m_B = 40 \,kg)$:
$f_{s,max} = \mu_s m_A g = 0.60 \times 10 \times 9.8 = 58.8 \,N$.
$2$. Compare the applied force $(F = 100 \,N)$ with the limiting friction:
Since $F > f_{s,max}$ $(100 \,N > 58.8 \,N)$,the block will slip over the slab.
$3$. Calculate the kinetic friction force acting between the block and the slab:
$f_k = \mu_k m_A g = 0.40 \times 10 \times 9.8 = 39.2 \,N$.
$4$. This kinetic friction force $f_k$ acts on the slab in the direction of the applied force,causing it to accelerate:
$F_{net, slab} = f_k = 39.2 \,N$.
$5$. Calculate the acceleration of the slab $(a_B)$:
$a_B = \frac{F_{net, slab}}{m_B} = \frac{39.2}{40} = 0.98 \,m/s^2$.

(A) Let $m_A = 2 \ kg$ and $m_B = 8 \ kg$. The force applied is $F = 25 \ N$.
First,check if the system moves. The maximum static friction between block $B$ and the ground is $f_{g,max} = \mu_B (m_A + m_B)g = 0.5 \times (2 + 8) \times 10 = 50 \ N$.
Since the applied force $F = 25 \ N$ is less than the maximum static friction $f_{g,max} = 50 \ N$,the entire system remains at rest.
Because the system is in equilibrium and no external force is trying to move block $A$ relative to block $B$,the frictional force between the two blocks is $0 \ N$.

(B) To move block $B$,the applied force $P$ must overcome the frictional forces acting on it.
$1$. The friction between block $A$ and block $B$ $(F_{AB})$ acts on the top surface of block $B$ in the direction opposite to the motion. Since block $A$ is stationary,$F_{AB} = \mu_{AB} \cdot N_A = \mu_{AB} \cdot m_A \cdot g = 0.25 \times 100 \times 10 = 250 \, N$.
$2$. The friction between block $B$ and the ground $(F_{BS})$ acts on the bottom surface of block $B$. The normal force on the ground is the sum of the weights of both blocks: $N_{ground} = (m_A + m_B)g = (100 + 200) \times 10 = 3000 \, N$. Thus,$F_{BS} = \mu_{BS} \cdot N_{ground} = (1/3) \times 3000 = 1000 \, N$.
$3$. The total force $P$ required is the sum of these two frictional forces: $P = F_{AB} + F_{BS} = 250 + 1000 = 1250 \, N$.

(C) Let the masses be $m_3 = 3 \ kg$,$m_2 = 2 \ kg$,and $m_1 = 1 \ kg$. The acceleration due to gravity is $g = 10 \ m/s^2$.
$1$. Minimum force required for slipping to occur between $3 \ kg$ and $2 \ kg$ block:
$F_{s1} = \mu_1 \cdot m_3 \cdot g = 0.5 \times 3 \times 10 = 15 \ N$.
$2$. Minimum force required for slipping to occur between $2 \ kg$ and $1 \ kg$ block:
$F_{s2} = \mu_2 \cdot (m_3 + m_2) \cdot g = 0.3 \times (3 + 2) \times 10 = 0.3 \times 50 = 15 \ N$.
$3$. Minimum force required for slipping to occur between $1 \ kg$ block and ground:
$F_{s3} = \mu_3 \cdot (m_3 + m_2 + m_1) \cdot g = 0.1 \times (3 + 2 + 1) \times 10 = 0.1 \times 60 = 6 \ N$.
Since the minimum force required to initiate slipping is least for the surface between the $1 \ kg$ block and the ground $(6 \ N)$,slipping will start there first.

(D) $1$. Analyze the friction between the $2 \ kg$ and $3 \ kg$ blocks: The maximum static friction force is $f_{max1} = \mu_1 N_1 = 0.2 \times 2 \times 10 = 4 \ N$. The applied force on the $2 \ kg$ block is $5 \ N$. Since $5 \ N > 4 \ N$,the blocks will slip relative to each other. The kinetic friction force is $f_k = 4 \ N$.
$2$. Analyze the friction between the $3 \ kg$ block and the ground: The normal force is $N_2 = (2 + 3) \times 10 = 50 \ N$. The maximum static friction force is $f_{max2} = \mu_2 N_2 = 0.1 \times 50 = 5 \ N$.
$3$. Calculate acceleration of the $2 \ kg$ block $(a_1)$: The net force is $F_{net1} = 5 \ N - 4 \ N = 1 \ N$. Thus,$a_1 = 1 \ N / 2 \ kg = 0.5 \ m/s^2$.
$4$. Calculate acceleration of the $3 \ kg$ block $(a_2)$: The forces acting on the $3 \ kg$ block are the applied $10 \ N$ force,the kinetic friction from the $2 \ kg$ block $(4 \ N)$,and the friction from the ground $(5 \ N)$. The net force is $F_{net2} = 10 \ N + 4 \ N - 5 \ N = 9 \ N$. Thus,$a_2 = 9 \ N / 3 \ kg = 3 \ m/s^2$.

$(A)$ Let $m_1 = 10 \text{ kg}$ (top block) and $m_2 = 5 \text{ kg}$ (bottom block). The coefficient of friction between the blocks is $\mu_1 = 0.1$ and between the bottom block and the ground is $\mu_2 = 0.3$. Taking $g = 10 \text{ m/s}^2$.
Maximum static friction between the blocks: $f_{1,max} = \mu_1 m_1 g = 0.1 \times 10 \times 10 = 10 \text{ N}$.
Maximum static friction between the bottom block and the ground: $f_{2,max} = \mu_2 (m_1 + m_2) g = 0.3 \times (10 + 5) \times 10 = 0.3 \times 150 = 45 \text{ N}$.
When an external force $F = 2 \text{ N}$ is applied to the top block, the system will only move if $F > f_{2,max}$. Since $F = 2 \text{ N}$ is less than both $f_{1,max}$ and $f_{2,max}$, the entire system remains at rest.
For the system to be in equilibrium, the frictional force from the ground must balance the applied force $F$. Therefore, the frictional force between the $5 \text{ kg}$ block and the ground is equal to the applied force $F = 2 \text{ N}$.

(A) Let $m_1 = 10 \text{ kg}$ and $m_2 = 5 \text{ kg}$. The coefficient of friction between the blocks is $\mu_1 = 0.1$ and between the $5 \text{ kg}$ block and the ground is $\mu_2 = 0.3$.
First,calculate the maximum static friction between the $10 \text{ kg}$ and $5 \text{ kg}$ blocks: $f_{max1} = \mu_1 m_1 g = 0.1 \times 10 \times 10 = 10 \text{ N}$.
Next,calculate the maximum static friction between the $5 \text{ kg}$ block and the ground: $f_{max2} = \mu_2 (m_1 + m_2) g = 0.3 \times (10 + 5) \times 10 = 0.3 \times 150 = 45 \text{ N}$.
Since the applied force $F = 2 \text{ N}$ is less than $f_{max1} = 10 \text{ N}$,the $10 \text{ kg}$ block will not slide over the $5 \text{ kg}$ block.
Therefore,the frictional force between the blocks is equal to the applied force $F$,which is $2 \text{ N}$.

(A) Let $m_1 = 10 \ kg$ (upper block) and $m_2 = 5 \ kg$ (lower block). The coefficient of friction between the blocks is $\mu_1 = 0.1$ and between the lower block and the ground is $\mu_2 = 0.3$.
For the system to remain at rest,the force $F$ must be balanced by the static friction force between the lower block and the ground.
The maximum static friction force between the lower block and the ground is $f_{max, ground} = \mu_2 (m_1 + m_2) g = 0.3 \times (10 + 5) \times 10 = 0.3 \times 150 = 45 \ N$.
However,we must also ensure the upper block does not slide over the lower block. The maximum static friction force between the two blocks is $f_{max, block} = \mu_1 m_1 g = 0.1 \times 10 \times 10 = 10 \ N$.
If $F > 10 \ N$,the upper block will start moving relative to the lower block.
Therefore,the maximum force $F$ that can be applied without causing any motion is $10 \ N$.

(C) Let the $10\, kg$ block be $m_1$ and the $5\, kg$ block be $m_2$. The coefficient of friction between $m_1$ and $m_2$ is $\mu_1 = 0.1$,and between $m_2$ and the ground is $\mu_2 = 0.3$.
The maximum static friction force between $m_1$ and $m_2$ is $f_{1,max} = \mu_1 m_1 g = 0.1 \times 10 \times 10 = 10\, N$.
The maximum static friction force between $m_2$ and the ground is $f_{2,max} = \mu_2 (m_1 + m_2) g = 0.3 \times (10 + 5) \times 10 = 0.3 \times 150 = 45\, N$.
For the $5\, kg$ block $(m_2)$ to move,the force applied by the $10\, kg$ block on it (which is the friction $f_1$) must exceed the maximum static friction force from the ground $(f_{2,max} = 45\, N)$.
Since the maximum force that $m_1$ can exert on $m_2$ is $f_{1,max} = 10\, N$,and $10\, N < 45\, N$,the $5\, kg$ block will never move regardless of the force $F$ applied to the $10\, kg$ block.
Therefore,the acceleration of the $5\, kg$ block is $0\, m/s^2$.

(A) Given: Mass of upper block $m_1 = 10\, kg$,mass of lower block $m_2 = 5\, kg$,coefficient of friction between blocks $\mu_1 = 0.1$,coefficient of friction between lower block and ground $\mu_2 = 0.3$,and applied force $F = 30\, N$.
First,calculate the maximum static friction force between the two blocks: $f_{max1} = \mu_1 m_1 g = 0.1 \times 10 \times 10 = 10\, N$.
Next,calculate the maximum static friction force between the lower block and the ground: $f_{max2} = \mu_2 (m_1 + m_2) g = 0.3 \times (10 + 5) \times 10 = 0.3 \times 150 = 45\, N$.
Since the applied force $F = 30\, N$ is less than the force required to move the entire system $(f_{max2} = 45\, N)$,the lower block will not move.
Now,check if the upper block moves relative to the lower block: The applied force $F = 30\, N$ is greater than the maximum friction $f_{max1} = 10\, N$ between the blocks.
Therefore,the upper block will slide over the lower block. The net force on the $10\, kg$ block is $F_{net} = F - f_{max1} = 30 - 10 = 20\, N$.
The acceleration of the $10\, kg$ block is $a = \frac{F_{net}}{m_1} = \frac{20}{10} = 2\, m/s^2$.

(C) Let $m_1 = 4 \, kg$ (top block $P$) and $m_2 = 5 \, kg$ (bottom block $Q$).
Coefficient of friction between $P$ and $Q$ is $\mu_1 = 0.2$. Coefficient of friction between $Q$ and ground is $\mu_2 = 0.1$.
Maximum static friction between $P$ and $Q$ is $f_{1,max} = \mu_1 m_1 g = 0.2 \times 4 \times 10 = 8 \, N$.
Maximum static friction between $Q$ and ground is $f_{2,max} = \mu_2 (m_1 + m_2) g = 0.1 \times (4 + 5) \times 10 = 9 \, N$.
For the system to move,the applied force $F$ must exceed $f_{2,max} = 9 \, N$. Thus,minimum force is $9 \, N$. Statement $A$ is incorrect.
If $F = 4 \, N < 9 \, N$,the system remains at rest. The friction from the ground balances $F$,so $f_2 = 4 \, N$. There is no tendency for $P$ to move relative to $Q$,so $f_1 = 0$. Statement $B$ is incorrect.
Maximum acceleration of $P$ is $a_{max} = f_{1,max} / m_1 = 8 / 4 = 2 \, m/s^2$. Statement $C$ is correct.
Slipping starts when the acceleration of $Q$ exceeds $2 \, m/s^2$. For the whole system,$F - f_{2,max} = (m_1 + m_2) a$. At $a = 2 \, m/s^2$,$F - 9 = (9) \times 2 \Rightarrow F = 27 \, N$. Statement $D$ is incorrect.

(D) Let $a$ be the acceleration of the system consisting of blocks $M$ and $m_0$ moving horizontally,and block $m$ moving vertically downwards. Since the string is inextensible,all blocks have the same magnitude of acceleration $a$.
For the system $(M + m_0)$,the equation of motion is: $T = (M + m_0)a$.
For the block $m$,the equation of motion is: $mg - T = ma$.
Adding these two equations: $mg = (M + m_0 + m)a$,so $a = \frac{mg}{M + m_0 + m}$.
For the block $m_0$ to remain stationary with respect to $M$,the static friction $f$ must provide the necessary pseudo force: $f = m_0 a$.
The maximum static friction is $f_{max} = \mu N = \mu m_0 g$.
For $m_0$ to not slide,we require $f \le f_{max}$,so $m_0 a \le \mu m_0 g$.
Substituting $a$: $m_0 \left( \frac{mg}{M + m_0 + m} \right) \le \mu m_0 g$.
Thus,$\mu \ge \frac{m}{M + m_0 + m}$.
Since this expression is not among the given options,the correct choice is $(D)$.

(A) Let the acceleration of the system be $a$. Since all three blocks move together,they share the same acceleration $a$. The total mass of the system is $(M + m_0 + m)$. The driving force is the weight of block $m$,which is $mg$. Applying Newton's second law to the whole system: $mg = (M + m_0 + m)a$,so $a = \frac{mg}{M + m_0 + m}$.
Now,consider the block $m_0$. The only horizontal force acting on $m_0$ is the static friction $f$ exerted by block $M$. For $m_0$ to move with acceleration $a$,we must have $f = m_0 a$.
The maximum value of static friction is $f_{max} = \mu N = \mu m_0 g$,where $N = m_0 g$ is the normal force between $m_0$ and $M$.
For the blocks to move together,the required friction must not exceed the maximum static friction: $f \le f_{max}$.
Substituting the expressions: $m_0 a \le \mu m_0 g$.
$a \le \mu g$.
$\frac{mg}{M + m_0 + m} \le \mu g$.
$\mu \ge \frac{m}{M + m_0 + m}$.
Thus,the minimum value of $\mu$ is $\frac{m}{M + m_0 + m}$.

(A) Let the acceleration of the system be $a = F / (M + m + M_0)$.
For block $m$ to be stationary with respect to $M_0$,the normal force $N$ exerted by $M_0$ on $m$ must provide the acceleration $a$. Thus,$N = ma$.
The vertical forces on $m$ are gravity $mg$ downwards and friction $f$ upwards. For equilibrium,$f = mg$.
Since $f \le \mu N$,we have $mg \le \mu (ma)$,which implies $a \ge g / \mu$.
If $F = 0$,then $a = 0$,so $f = mg$ cannot be balanced,meaning the blocks cannot remain stationary. Thus,$(A)$ is correct.
For $f = 0$,we need $mg = 0$,which is impossible. Thus,$(C)$ is incorrect.
For a range of $F$ such that $a \ge g / \mu$,the blocks can remain stationary,not just for one unique value. Thus,$(B)$ is incorrect.
Therefore,only statement $(A)$ is correct.

(D) Let the acceleration of the system be $a$. Since $M$ and $m$ are stationary with respect to $M_0$,the whole system moves with acceleration $a = \frac{F}{M_0 + M + m}$.
For block $M$: The pseudo force $Ma$ acts to the right. The tension $T$ in the string pulls it to the right. So,$T = Ma$.
For block $m$: The weight $mg$ acts downwards. The friction force $f = \mu N$ acts upwards,where $N$ is the normal force. The pseudo force $ma$ acts horizontally on $m$ against the vertical surface,so $N = ma$.
For $m$ to be stationary,the vertical forces must balance: $f = mg \implies \mu N = mg \implies \mu(ma) = mg \implies a = \frac{g}{\mu}$.
Substituting $a$ into the system equation: $\frac{F}{M_0 + M + m} = \frac{g}{\mu} \implies F = (M_0 + M + m) \frac{g}{\mu}$.
Since this result is not among the options,the correct choice is $D$.

(D) Let the mass of block $D$ be $M$. The blocks $A, B, C$ move together with acceleration $a$. The maximum acceleration $a_{max}$ that blocks $B$ and $C$ can have without slipping is determined by the friction force. For block $C$,the maximum friction force is $f_{max} = \mu mg$. Thus,$a_{max} = \frac{f_{max}}{m} = \mu g$.
For the system of blocks $(A+B+C)$,the total mass is $3m$. The tension $T$ in the string provides the acceleration:
$T = (3m)a = (3m)(\mu g) = 3m\mu g$.
For block $D$ of mass $M$ moving downwards:
$Mg - T = Ma$
$Mg - 3m\mu g = M(\mu g)$
$Mg - M\mu g = 3m\mu g$
$Mg(1 - \mu) = 3m\mu g$
$M = \frac{3m\mu}{1 - \mu}$.

(D) The kinetic friction force acting between the blocks is $f_k = \mu N = 0.3 \times (2 \times 10) = 6\ N$.
For the $2\ kg$ block,the net force is $F_{net,1} = 10 - 6 = 4\ N$. Its acceleration is $a_1 = \frac{4}{2} = 2\ m s^{-2}$.
For the $8\ kg$ block,the only horizontal force is the friction force $f_k = 6\ N$ acting in the forward direction. Its acceleration is $a_2 = \frac{6}{8} = 0.75\ m s^{-2}$.
The relative acceleration of the $2\ kg$ block with respect to the $8\ kg$ block is $a_{rel} = a_1 - a_2 = 2 - 0.75 = 1.25\ m s^{-2}$.
Using the equation of motion $s = ut + \frac{1}{2} a_{rel} t^2$,where $s = 3.0\ m$ and $u = 0$:
$3.0 = 0 + \frac{1}{2} \times 1.25 \times t^2$
$t^2 = \frac{3.0 \times 2}{1.25} = \frac{6}{1.25} = 4.8$
$t = \sqrt{4.8} \approx 2.19\ s$.

(C) Let $m_1 = 10 \ kg$ and $m_2 = 20 \ kg$. The coefficients of friction are $\mu_1 = 0.3$ and $\mu_2$.
At the limiting equilibrium,the total gravitational force component along the incline must be equal to the sum of the maximum static frictional forces on both blocks.
$(m_1 + m_2) g \sin \theta = (\mu_1 m_1 g \cos \theta + \mu_2 m_2 g \cos \theta)$
$(10 + 20) g \sin \theta = (0.3 \times 10 \times g \cos \theta + \mu_2 \times 20 \times g \cos \theta)$
Dividing by $g \cos \theta$,we get:
$30 \tan \theta = 3 + 20 \mu_2$
Given $\tan \theta = 1/2$,so $30(1/2) = 3 + 20 \mu_2$
$15 = 3 + 20 \mu_2$
$12 = 20 \mu_2$
$\mu_2 = 12/20 = 0.6$

(B) The frictional force between the blocks is $f = \mu N = 0.5 \times 4 \times 10 = 20\ N$.
For the $4\ kg$ block (moving right at $12\ m/s$): $a_2 = -f/m_2 = -20/4 = -5\ m/s^2$.
For the $2\ kg$ block (moving left at $6\ m/s$): $a_1 = +f/m_1 = +20/2 = +10\ m/s^2$.
Relative motion stops when $v_1 = v_2$:
$-6 + 10t = 12 - 5t$
$15t = 18 \implies t = 1.2\ s$. (Statement $ii$ is correct).
Common velocity $v = 12 - 5(1.2) = 6\ m/s$ (towards the right). (Statement $iii$ is incorrect).
Displacement of $4\ kg$ block: $s = ut + 0.5at^2 = 12(1.2) + 0.5(-5)(1.2)^2 = 14.4 - 3.6 = 10.8\ m$. (Statement $iv$ is correct).
Thus,statements $ii$ and $iv$ are correct.

(C) For the block $m$ to move with the wedge $M$, the vertical forces must balance. The pseudo force acting on $m$ is $F_p = ma$ in the backward direction. This pseudo force acts as the normal force $N$ between $m$ and $M$, so $N = ma = 1 \times 20 = 20\, N$.
The maximum static friction force is $f_{max} = \mu N = 0.6 \times 20 = 12\, N$.
The weight of the block is $mg = 1 \times 10 = 10\, N$.
Since $f_{max} > mg$, the block $m$ will not slide down. The static friction force $f$ balances the weight of the block, so $f = mg = 10\, N$.
As long as the block does not slide, the friction force $f$ remains equal to $mg$, which is constant, regardless of small changes in acceleration $a$, provided $f_{max} \ge mg$ (i.e., $\mu ma \ge mg \implies a \ge g/\mu = 10/0.6 = 16.67\, m/s^2$).
Since the current acceleration $20\, m/s^2 > 16.67\, m/s^2$, the friction force is constant at $10\, N$.

(C) $1$. Calculate the friction forces:
- Friction between $A$ and $B$: $f_1 = \mu_1 m_A g = 0.4 \times 5 \times 10 = 20 \ N$.
- Friction between $B$ and ground: $f_2 = \mu_2 (m_A + m_B) g = 0.2 \times (5 + 4) \times 10 = 0.2 \times 90 = 18 \ N$.
$2$. Calculate accelerations:
- For block $A$: $a_A = f_1 / m_A = 20 / 5 = 4 \ m/s^2$ (towards the right).
- For block $B$: $F - f_1 - f_2 = m_B a_B \Rightarrow 62 - 20 - 18 = 4 a_B \Rightarrow 24 = 4 a_B \Rightarrow a_B = 6 \ m/s^2$ (towards the right).
$3$. Relative motion:
- Relative acceleration $a_{B/A} = a_B - a_A = 6 - 4 = 2 \ m/s^2$ (towards the right).
- Using $s = ut + 1/2 a t^2$,where $s = 16 \ m$,$u = 0$,$a = 2 \ m/s^2$:
- $16 = 0 + 1/2 \times 2 \times t^2 \Rightarrow 16 = t^2 \Rightarrow t = 4 \ \text{sec}$.

(B) Let the masses be $m_1 = 1 \ kg$ and $m_2 = 2 \ kg$. The applied force is $F = 0.5 \ N$.
Assuming both blocks move together with acceleration $a$,the equation of motion for the system is $F = (m_1 + m_2)a$.
$0.5 = (1 + 2)a \Rightarrow 0.5 = 3a \Rightarrow a = \frac{0.5}{3} = \frac{1}{6} \ m/s^2$.
The friction force $f$ acting on the lower block $(m_2)$ is $f = m_2 a = 2 \times \frac{1}{6} = \frac{1}{3} \ N$.
The maximum static friction force is $f_{max} = \mu_s m_1 g = 0.1 \times 1 \times 10 = 1 \ N$.
Since $f = \frac{1}{3} \ N < f_{max} = 1 \ N$,the assumption that they move together is correct.
The lower block exerts a friction force $f$ in the backward direction on the upper block.
Work done by the lower block on the upper block $W = -f \times d$,where $d = 3 \ m$.
$W = -(\frac{1}{3}) \times 3 = -1 \ J$.

(D) For both blocks to move together,block $A$ must move with the same acceleration $a$ as block $B$. The only force causing acceleration on block $A$ is the static friction force $f_s$ from block $B$.
The maximum static friction force between $A$ and $B$ is $f_{s,max} = \mu_2 m_A g = 0.3 \times 20 \times 10 = 60 \text{ N}$.
The maximum acceleration $a_{max}$ of block $A$ is $a_{max} = \frac{f_{s,max}}{m_A} = \frac{60}{20} = 3 \text{ m/s}^2$.
Now,consider the system of both blocks $(A+B)$ moving together with acceleration $a_{max} = 3 \text{ m/s}^2$. The external force $F$ must overcome the kinetic friction $f_k$ between block $B$ and the floor.
The normal force on the floor is $N = (m_A + m_B)g = (20 + 40) \times 10 = 600 \text{ N}$.
The kinetic friction force is $f_k = \mu_1 N = 0.2 \times 600 = 120 \text{ N}$.
Applying Newton's second law to the system $(A+B)$:
$F - f_k = (m_A + m_B) a_{max}$
$F - 120 = (20 + 40) \times 3$
$F - 120 = 60 \times 3$
$F - 120 = 180$
$F = 300 \text{ N}$.

(A) $1$. First,calculate the maximum static friction force between the blocks: $f_{s,max} = \mu_s N = \mu_s m_B g = 0.4 \times 5 \times 10 = 20\, N$.
$2$. If both blocks move together with acceleration $a$,then $F = (m_A + m_B) a$. Thus,$175 = (30 + 5) a \Rightarrow a = 175 / 35 = 5\, m/s^2$.
$3$. The force required to accelerate block $A$ at $5\, m/s^2$ is $F_A = m_A a = 30 \times 5 = 150\, N$. Since $150\, N > f_{s,max} (20\, N)$,the blocks will slip.
$4$. Since they slip,the kinetic friction force $f_k = \mu_k m_B g = 0.3 \times 5 \times 10 = 15\, N$ acts on block $A$.
$5$. The acceleration of block $A$ is $a_A = f_k / m_A = 15 / 30 = 0.5\, m/s^2$.

(C) Let the acceleration of the system be $a$. For the $4 \ kg$ block to move with the $8 \ kg$ block without slipping,the maximum frictional force must provide the necessary acceleration.
The limiting friction on the $4 \ kg$ block is $f_{max} = \mu N = \mu mg = 0.4 \times 4 \times 10 = 16 \ N$.
This frictional force provides the acceleration to the $4 \ kg$ block: $f_{max} = m_1 a \Rightarrow 16 = 4a \Rightarrow a = 4 \ m/s^2$.
Now,considering the whole system of $(4 \ kg + 8 \ kg)$ moving with acceleration $a$ under force $F$:
$F = (m_1 + m_2) a = (4 + 8) \times 4 = 12 \times 4 = 48 \ N$.
Thus,the maximum force $F$ is $48 \ N$.

(A) $1$. Assume both blocks $A$ and $B$ move together with a common acceleration $a$. The total mass of the system is $M = m_A + m_B = 1 + 2 = 3\,kg$.
$2$. The external force applied is $F = 5\,N$. The acceleration of the system is $a = \frac{F}{M} = \frac{5}{3}\,m/s^2$.
$3$. For block $A$ to move with this acceleration,the required force is $f = m_A \cdot a = 1 \cdot \frac{5}{3} = \frac{5}{3}\,N$.
$4$. The maximum static friction force between $A$ and $B$ is $f_{max} = \mu \cdot m_A \cdot g = 0.2 \cdot 1 \cdot 10 = 2\,N$.
$5$. Since the required force $f = \frac{5}{3}\,N \approx 1.67\,N$ is less than $f_{max} = 2\,N$,the blocks move together.
$6$. Therefore,the frictional force exerted by $A$ on $B$ is equal to the force required to accelerate $A$,which is $\frac{5}{3}\,N$.

(C) Let the coefficient of friction between blocks $A$ and $B$ be $\mu$. The maximum static friction force between $A$ and $B$ is $f_{max} = \mu m_A g = \mu \times 4 \times 10 = 40\mu$.
When a force $F_A = 12\, N$ is applied on $A$,the acceleration of the system is $a = \frac{F_A}{m_A + m_B} = \frac{12}{4 + 5} = \frac{12}{9} = \frac{4}{3}\, m/s^2$.
For block $A$ to move with $B$,the friction force $f$ must provide the acceleration to $A$: $f = m_A a = 4 \times \frac{4}{3} = \frac{16}{3}\, N$.
Since this is the limiting case,$f = f_{max} \Rightarrow 40\mu = \frac{16}{3} \Rightarrow \mu = \frac{16}{120} = \frac{2}{15}$.
Now,let a maximum force $F_B$ be applied to block $B$. The acceleration of the system is $a' = \frac{F_B}{m_A + m_B} = \frac{F_B}{9}$.
For block $A$ to move with $B$,the friction force $f$ must provide the acceleration to $A$: $f = m_A a' = 4 \times \frac{F_B}{9}$.
Since $f \le f_{max}$,we have $4 \times \frac{F_B}{9} \le 40 \times \frac{2}{15}$.
$F_B \le 40 \times \frac{2}{15} \times \frac{9}{4} = 10 \times 2 \times \frac{3}{5} \times 3 = 36\, N$. Wait,re-calculating: $F_B \le 40 \times \frac{2}{15} \times \frac{9}{4} = 10 \times 2 \times \frac{9}{15} = 20 \times 0.6 = 12\, N$ is incorrect. Let's re-evaluate: $f_{max} = 40 \times \frac{2}{15} = \frac{16}{3}$. $4 \times \frac{F_B}{9} = \frac{16}{3} \Rightarrow F_B = \frac{16}{3} \times \frac{9}{4} = 4 \times 3 = 12\, N$. Actually,the question implies $F_B$ is applied to $B$ only. $F_B - f = m_B a' \Rightarrow F_B - \frac{16}{3} = 5 \times \frac{F_B}{9} \Rightarrow F_B(1 - 5/9) = 16/3 \Rightarrow F_B(4/9) = 16/3 \Rightarrow F_B = 16/3 \times 9/4 = 12\, N$. Given the options,let's re-check the initial $\mu$ calculation. $12 - f = 4a$ and $f = 5a$. $12 = 9a \Rightarrow a = 4/3$. $f = 5(4/3) = 20/3$. $f_{max} = \mu(4)(10) = 40\mu = 20/3 \Rightarrow \mu = 1/6$. Now for $B$: $F_B - f = 5a'$ and $f = 4a'$. $F_B = 9a'$. Max $f = 40(1/6) = 20/3$. $4a' = 20/3 \Rightarrow a' = 5/3$. $F_B = 9(5/3) = 15\, N$.

(B) For block $A$ not to slide over $B$,the maximum acceleration $a$ of the system is limited by the friction between $A$ and $B$.
The maximum frictional force on $A$ is $f_{max} = \mu_1 m_A g = 0.2 \times 1 \times 10 = 2\,N$.
This friction provides the maximum acceleration to block $A$: $a_{max} = \frac{f_{max}}{m_A} = \frac{2}{1} = 2\,m/s^2$.
Now,consider the whole system $(A+B)$ moving with acceleration $a_{max}$. The external force $F$ must overcome the friction between $B$ and the table surface $(f_{table})$ and provide the required acceleration to both blocks.
The friction between $B$ and the table is $f_{table} = \mu_2 (m_A + m_B) g = 0.2 \times (1 + 3) \times 10 = 0.2 \times 4 \times 10 = 8\,N$.
Applying Newton's second law to the system $(A+B)$: $F - f_{table} = (m_A + m_B) a_{max}$.
$F - 8 = (1 + 3) \times 2$.
$F - 8 = 4 \times 2 = 8$.
$F = 8 + 8 = 16\,N$.

(D) Given: Mass of block $A$,$m_{A} = \frac{m}{2}$. Mass of block $B$,$m_{B} = m$.
Coefficient of friction between $A$ and $B$,$\mu_{A} = 0.2$. Coefficient of friction between $B$ and the floor,$\mu_{B} = 0.1$.
For the blocks to move together,the maximum acceleration $a$ is provided by the friction between $A$ and $B$.
$f_{max} = \mu_{A} m_{A} g = m_{A} a$
$a = \mu_{A} g = 0.2 g$.
Now,considering both blocks as a single system,the total mass is $M = m_{A} + m_{B} = \frac{m}{2} + m = \frac{3m}{2}$.
The force $F$ must overcome the friction between $B$ and the floor,and provide acceleration $a$ to the system.
$F - \mu_{B} (m_{A} + m_{B}) g = (m_{A} + m_{B}) a$
$F = (m_{A} + m_{B}) a + \mu_{B} (m_{A} + m_{B}) g$
$F = (\frac{3m}{2})(0.2 g) + (0.1)(\frac{3m}{2}) g$
$F = 0.3 mg + 0.15 mg = 0.45 mg$.
Thus,the maximum force is $0.45 mg$.

(B) The maximum static frictional force (limiting friction) is given by $f_{max} = \mu_s N = \mu_s mg$.
Given $\mu_s = 0.6$,$m = 1\, kg$,and $g = 9.8\, m/s^2$.
$f_{max} = 0.6 \times 1 \times 9.8 = 5.88\, N$.
The pseudo force acting on the block due to the acceleration of the truck is $F_{pseudo} = ma = 1 \times 5 = 5\, N$.
Since the applied pseudo force $(5\, N)$ is less than the limiting friction $(5.88\, N)$,the block does not slide relative to the truck.
Therefore,the static frictional force adjusts itself to balance the pseudo force.
Thus,the frictional force acting on the block is $5\, N$.

(C) To move block $B$,the applied force $F$ must overcome the frictional forces acting on it.
$1$. Friction between block $A$ and block $B$ $(f_{AB})$: This is the limiting friction acting on the top surface of block $B$ as it moves to the right. Since block $A$ is tied to the wall,it remains stationary. The normal force on $A$ is $N_A = m_A g = 100 \times 10 = 1000\, N$. Thus,$f_{AB} = \mu_{AB} N_A = 0.2 \times 1000 = 200\, N$.
$2$. Friction between block $B$ and the ground $(f_{BS})$: This is the limiting friction acting on the bottom surface of block $B$. The normal force on the ground is $N_G = (m_A + m_B) g = (100 + 200) \times 10 = 3000\, N$. Thus,$f_{BS} = \mu_{BS} N_G = 0.3 \times 3000 = 900\, N$.
$3$. Total force $F$ required: The force $F$ must overcome both frictional forces.
$F = f_{AB} + f_{BS} = 200\, N + 900\, N = 1100\, N$.

(C) The maximum frictional force $f_{max}$ acting on block $B$ is provided by the static friction between $A$ and $B$.
$f_{max} = \mu m_B g = 0.4 \times 3 \times 10 = 12 \, N$.
This frictional force provides the maximum acceleration $a_{max}$ to block $B$ so that it does not slip over $A$.
$a_{max} = \frac{f_{max}}{m_B} = \frac{12}{3} = 4 \, m/s^2$.
For both blocks to move together without separation,the entire system must move with this maximum acceleration $a_{max}$.
Applying Newton's second law to the combined system of blocks $A$ and $B$:
$F = (m_A + m_B) a_{max} = (6 + 3) \times 4 = 9 \times 4 = 36 \, N$.

(C) For the blocks to move together without separation,the maximum acceleration $a_{max}$ is determined by the limiting friction between block $A$ and block $B$.
The limiting frictional force $f_{max}$ acting on block $B$ is given by $f_{max} = \mu m_B g$.
Substituting the values,$f_{max} = 0.4 \times 3 \times 10 = 12\, N$.
This frictional force provides the necessary acceleration to block $B$. Thus,$f_{max} = m_B a_{max}$.
$12 = 3 \times a_{max} \implies a_{max} = 4\, m/s^2$.
Now,considering both blocks as a single system,the total force $F$ required to move them with acceleration $a_{max}$ is $F = (m_A + m_B) a_{max}$.
$F = (6 + 3) \times 4 = 9 \times 4 = 36\, N$.

(A) For the block of mass $m$ to move along with the block of mass $M$,the frictional force between them must provide the necessary acceleration to the block $m$.
The maximum static frictional force acting on the block $m$ is given by $f_{\max} = \mu N$,where $N$ is the normal force.
Since the block $m$ is on top of $M$,the normal force $N = mg$.
Therefore,the maximum frictional force is $f_{\max} = \mu mg$.
According to Newton's second law,the maximum acceleration $a_{\max}$ that can be provided to the block $m$ by this frictional force is $f_{\max} = m a_{\max}$.
Substituting the value of $f_{\max}$,we get $\mu mg = m a_{\max}$.
Solving for $a_{\max}$,we find $a_{\max} = \mu g$.

(A) For block $A$ to move with block $B$ without any relative motion,the maximum acceleration $a_{max}$ of the system is provided by the friction force between $A$ and $B$.
The limiting friction force on block $A$ is $f_{max} = \mu N_A = \mu m_A g$.
Using Newton's second law for block $A$,$f_{max} = m_A a_{max}$.
Therefore,$m_A a_{max} = \mu m_A g$,which gives $a_{max} = \mu g$.
Given $\mu = 0.60$ and $g = 10\,m/s^2$,we have $a_{max} = 0.60 \times 10 = 6\,m/s^2$.
Now,considering both blocks $A$ and $B$ as a single system of mass $(m_A + m_B)$,the maximum force $F_{max}$ that can be applied to $B$ is $F_{max} = (m_A + m_B) a_{max}$.
Substituting the values,$F_{max} = (2 + 5) \times 6 = 7 \times 6 = 42\,N$.

(B) Let $m_1 = 10 \, kg$ (upper block) and $m_2 = 5 \, kg$ (lower block).
The coefficient of friction between the blocks is $\mu_1 = 0.1$ and between the lower block and the ground is $\mu_2 = 0.3$.
The maximum static friction between the blocks is $f_{1 \max} = \mu_1 m_1 g = 0.1 \times 10 \times 10 = 10 \, N$.
The maximum static friction between the lower block and the ground is $f_{2 \max} = \mu_2 (m_1 + m_2) g = 0.3 \times (10 + 5) \times 10 = 0.3 \times 150 = 45 \, N$.
When an external force $F = 2 \, N$ is applied to the upper block,the force trying to move the system is $F = 2 \, N$.
Since $F < f_{1 \max}$ $(2 \, N < 10 \, N)$,the upper block does not move relative to the lower block.
Consequently,the entire system $(15 \, kg)$ remains at rest because the applied force $F = 2 \, N$ is also less than $f_{2 \max} = 45 \, N$.
Since the system is in equilibrium,the frictional force between the $5 \, kg$ block and the ground must balance the applied force $F$ to keep the system stationary.
Therefore,the frictional force between the $5 \, kg$ block and the ground is $2 \, N$.

(A) $1$. Identify the maximum static friction between the $10\,kg$ block and the $5\,kg$ block: $f_{max} = \mu_s N = 0.1 \times (10\,kg \times 10\,m/s^2) = 10\,N$.
$2$. The applied force $F = 2\,N$ is less than the maximum static friction $f_{max} = 10\,N$.
$3$. Since the applied force is insufficient to overcome the static friction,the blocks will not move relative to each other.
$4$. Therefore,the static frictional force acting between the blocks is equal to the applied force,which is $2\,N$.

(C) $1$. Calculate the limiting friction between the $10\,kg$ and $5\,kg$ blocks: $f_{1} = \mu_1 N_1 = 0.1 \times 10 \times 10 = 10\,N$.
$2$. The net force on the $10\,kg$ block is $F_{net} = F - f_1 = 30\,N - 10\,N = 20\,N$.
$3$. The acceleration of the $10\,kg$ block is $a = \frac{F_{net}}{m} = \frac{20\,N}{10\,kg} = 2\,m/s^2$.

(C) Mass of the box,$m = 40 \;kg$.
Coefficient of friction,$\mu = 0.15$.
Initial velocity,$u = 0$.
Acceleration of the truck,$a = 2 \;m s^{-2}$.
Distance of the box from the end of the truck,$s' = 5 \;m$.
In the frame of reference of the truck,the box experiences a pseudo force $F_p = ma$ in the backward direction.
$F_p = 40 \times 2 = 80 \;N$.
The limiting friction force acting in the forward direction is $f = \mu mg = 0.15 \times 40 \times 10 = 60 \;N$.
The net force on the box in the backward direction is $F_{net} = F_p - f = 80 - 60 = 20 \;N$.
The backward acceleration of the box relative to the truck is $a_{rel} = \frac{F_{net}}{m} = \frac{20}{40} = 0.5 \;m s^{-2}$.
Using the equation of motion $s' = u t + \frac{1}{2} a_{rel} t^2$:
$5 = 0 + \frac{1}{2} \times 0.5 \times t^2$
$5 = 0.25 \times t^2$
$t^2 = 20$
$t = \sqrt{20} \;s$.
The distance $s$ traveled by the truck in time $t$ is:
$s = ut + \frac{1}{2} a t^2$
$s = 0 + \frac{1}{2} \times 2 \times (\sqrt{20})^2$
$s = 20 \;m$.

(C) For the blocks to move together,the acceleration $a$ of the system must be such that the frictional force on the smaller block $m$ is sufficient to provide its acceleration.
The maximum frictional force available is $f_{\max} = \mu N = \mu mg$.
The maximum acceleration $a_{\max}$ that the block $m$ can have without slipping is $a_{\max} = \frac{f_{\max}}{m} = \mu g$.
Given $\mu = \frac{3}{7}$ and $g = 9.8\, m/s^2$,we have:
$a_{\max} = \frac{3}{7} \times 9.8 = 3 \times 1.4 = 4.2\, m/s^2$.
Now,for the entire system of mass $(M + m)$,the force $F$ is applied:
$F = (M + m) a_{\max}$.
Substituting the values:
$F = (4.5 + 0.5) \times 4.2 = 5 \times 4.2 = 21\, N$.
Thus,the maximum horizontal force is $21\, N$.

(D) Let the mass of the upper block be $m_1 = 1 \, kg$ and the lower block be $m_2 = 2 \, kg$. The total mass of the system is $M = m_1 + m_2 = 3 \, kg$.
For the blocks to move together,the upper block must move with the same acceleration $a$ as the lower block. The only force causing the upper block to accelerate is the static friction force $f_s$ between the two blocks.
The maximum static friction force is $f_{s,max} = \mu m_1 g = 0.5 \times 1 \times 10 = 5 \, N$.
This friction force provides the maximum acceleration to the $1 \, kg$ block: $f_{s,max} = m_1 a \Rightarrow 5 = 1 \times a \Rightarrow a = 5 \, m/s^2$.
Now,considering the whole system of $3 \, kg$ moving with acceleration $a$,the applied force $F$ is given by $F = (m_1 + m_2) a = 3 \times 5 = 15 \, N$.

(C) The maximum acceleration $a_{\max}$ that the block of mass $m$ can have without slipping is determined by the static friction force $f_{s, \max} = \mu_s N = \mu_s mg$.
Using Newton's second law for block $m$: $f_{s, \max} = m a_{\max} \implies \mu_s mg = m a_{\max} \implies a_{\max} = \mu_s g$.
Given $\mu_s = 0.5$ and $g = 9.8 \; m/s^2$,we have $a_{\max} = 0.5 \times 9.8 = 4.9 \; m/s^2$.
For the two blocks to move together,the entire system must move with an acceleration $a \le a_{\max}$.
Applying Newton's second law to the combined system of mass $(m + M)$:
$F = (m + M) a_{\max} = (2 + 8) \times 4.9 = 10 \times 4.9 = 49 \; N$.

(A) Let the acceleration of the system of blocks $X$ and $Y$ be $a$.
Since the blocks move together,the total mass is $M = m_X + m_Y = 8 \,kg + 2 \,kg = 10 \,kg$.
The acceleration of the system is given by $a = \frac{F}{M} = \frac{F}{10} \,ms^{-2}$.
For block $Y$ to not slip downwards,the upward frictional force $f$ must balance its weight $m_Y g$.
$f = m_Y g = 2 \times 10 = 20 \,N$.
The frictional force is also given by $f = \mu R$,where $R$ is the normal reaction between blocks $X$ and $Y$.
From the horizontal motion of block $Y$,the normal reaction $R$ provides the necessary acceleration $a$ to block $Y$:
$R = m_Y a = 2 \times \frac{F}{10} = \frac{F}{5}$.
Substituting the values into the friction equation:
$f = \mu R \Rightarrow 20 = 0.5 \times \frac{F}{5}$.
$20 = \frac{F}{10} \Rightarrow F = 200 \,N$.
Thus,the minimum value of $F$ is $200 \,N$.

(A) The limiting friction between the pen and the paper is $f_1 = \mu_1 mg$.
For the pen to start slipping,the force acting on the pen must be at least $f_1$. This force is provided by the friction from the paper,which is equal to the force of friction on the pen.
Thus,the acceleration $a$ of the pen is $a = \frac{f_1}{m} = \mu_1 g$.
For the pen to slip,the paper must accelerate at least with this acceleration $a = \mu_1 g$.
Now,consider the free body diagram of the paper of mass $M$. The forces acting on the paper are the applied force $F$,the friction from the pen $f_1$ (acting backwards),and the friction from the table $f_2$ (acting backwards).
The normal force on the table is $N = (M+m)g$,so $f_2 = \mu_2(M+m)g$.
Applying Newton's second law to the paper:
$F - f_1 - f_2 = Ma$
$F = Ma + f_1 + f_2$
Substituting the values:
$F = M(\mu_1 g) + \mu_1 mg + \mu_2(M+m)g$
$F = (M+m)\mu_1 g + (M+m)\mu_2 g$
$F = (M+m)(\mu_1 + \mu_2)g$.

(D) Let both blocks move together with acceleration $a$.
For block $M_2$,the only horizontal force providing acceleration is the friction force $f_r$.
The maximum static friction force is $f_{r,max} = \mu N = \mu M_2 g$.
Thus,the maximum acceleration $a_{max}$ of block $M_2$ is $a_{max} = \frac{f_{r,max}}{M_2} = \frac{\mu M_2 g}{M_2} = \mu g$.
For the entire system of both blocks $(M_1 + M_2)$ to move together with this maximum acceleration $a_{max}$,the force $F$ applied on $M_1$ must satisfy:
$F = (M_1 + M_2) a_{max}$
$F = (M_1 + M_2) \mu g$
$F = \mu(M_1 + M_2) g$
Therefore,the correct option is $(d)$.

(C) For no relative motion between the blocks,both blocks must move with the same acceleration $a$.
Consider the block of mass $m_2$. The only horizontal force acting on it is the friction force $f$ exerted by block $m_1$.
Thus,$f = m_2 a$.
The maximum value of friction is $f_{\text{max}} = \mu N = \mu m_1 g$.
Therefore,the maximum acceleration $a_{\text{max}}$ that block $m_2$ can have is given by $m_2 a_{\text{max}} = \mu m_1 g$,which implies $a_{\text{max}} = \mu \frac{m_1}{m_2} g$.
Now,consider the system of both blocks $(m_1 + m_2)$. The external force $F$ is applied to $m_1$.
For the whole system,$F = (m_1 + m_2) a$.
Substituting $a_{\text{max}}$,we get $F_{\text{max}} = (m_1 + m_2) \mu \frac{m_1}{m_2} g$.
$F_{\text{max}} = \mu m_1 g \left(\frac{m_1 + m_2}{m_2}\right) = \mu m_1 g \left(\frac{m_1}{m_2} + 1\right)$.

(C) The only horizontal force acting on block $A$ (mass $m_A = 5 \, kg$) is the static frictional force $f$ exerted by block $B$.
According to Newton's second law of motion,the net force on an object is equal to the product of its mass and acceleration $(F_{net} = m \cdot a)$.
Here,the frictional force $f$ provides the acceleration $a_A = 2 \, m/s^2$ to block $A$.
Therefore,$f = m_A \cdot a_A$.
Substituting the given values: $f = 5 \, kg \times 2 \, m/s^2 = 10 \, N$.
Thus,the frictional force applied by block $B$ on block $A$ is $10 \, N$.

(A) The block $A$ is placed on block $B$. $A$ force $F$ is applied on block $B$. The block $A$ moves due to the friction force acting between $A$ and $B$.
From the graph,the acceleration of block $A$ increases linearly until it reaches $4.9 \ m/s^2$. At this point,the block $A$ is on the verge of slipping relative to block $B$.
The maximum static friction force $f_{max}$ provides the maximum acceleration to block $A$:
$f_{max} = m_A a_{max} = \mu_k m_A g$
Where $m_A$ is the mass of block $A$,$a_{max} = 4.9 \ m/s^2$ is the maximum acceleration before slipping,and $g = 9.8 \ m/s^2$ is the acceleration due to gravity.
$a_{max} = \mu_k g$
$4.9 = \mu_k \times 9.8$
$\mu_k = \frac{4.9}{9.8} = 0.5$
Thus,the coefficient of kinetic friction is $0.5$.

(C) Let the force $F$ be applied to block $A$ $(5 \,kg)$.
Since there is no relative slipping,both blocks move with the same acceleration $a$.
The total mass of the system is $M = 5 \,kg + 3 \,kg = 8 \,kg$.
The acceleration of the system is $a = \frac{F}{M} = \frac{F}{8}$.
For the upper block $B$ $(3 \,kg)$,the only horizontal force causing acceleration is the static friction $f$ between $A$ and $B$.
Thus,$f = m_B \cdot a = 3 \cdot \left(\frac{F}{8}\right) = \frac{3F}{8}$.
The maximum value of static friction is $f_{max} = \mu \cdot N = \mu \cdot m_B \cdot g = 0.5 \cdot 3 \cdot 10 = 15 \,N$ (taking $g = 10 \,m/s^2$).
For no relative slipping,$f \leq f_{max}$,so $\frac{3F}{8} \leq 15$.
$3F \leq 120 \implies F \leq 40 \,N$.
Since $1 \,kg \,wt. = 10 \,N$,the maximum force in $kg \,wt.$ is $\frac{40}{10} = 4 \,kg \,wt.$