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

(A) The maximum static friction force between the block and the slab is given by $f_{\text{max}} = \mu N = \mu m_1 g$,where $m_1 = 10 \,kg$ and $\mu = 0.40$.
$f_{\text{max}} = 0.40 \times 10 \times 10 = 40 \,N$.
This friction force acts on the $40 \,kg$ slab in the direction of the applied force.
The acceleration of the slab is caused by this friction force.
Using Newton's second law for the slab $(m_2 = 40 \,kg)$:
$F_{\text{friction}} = m_2 a_{\text{slab}}$
$40 \,N = 40 \,kg \times a_{\text{slab}}$
$a_{\text{slab}} = 1.0 \,m/s^2$.

(B) $1$. Assume both blocks move together with a common acceleration $a$. The total mass of the system is $M = 4 \,kg + 2 \,kg = 6 \,kg$.
$2$. The external force applied is $F = 30 \,N$. The common acceleration is $a = F / M = 30 / 6 = 5 \,m/s^2$.
$3$. To check if they move together,we calculate the required frictional force $f$ on the $2 \,kg$ block: $f = m \cdot a = 2 \,kg \times 5 \,m/s^2 = 10 \,N$.
$4$. The limiting friction $f_L$ between the blocks is $f_L = \mu_s \cdot N = \mu_s \cdot (m \cdot g) = 0.8 \times 2 \times 10 = 16 \,N$.
$5$. Since the required frictional force $(10 \,N)$ is less than the limiting friction $(16 \,N)$,the blocks will move together with an acceleration of $5 \,m/s^2$.

(C) Let $a$ be the common acceleration of the system consisting of blocks $A$ and $B$.
The total force applied is $F$,and the total mass is $(M+m)$.
Using Newton's second law for the whole system: $F = (M+m)a$,which gives $a = \frac{F}{M+m}$.
Now,consider the free body diagram of block $B$ of mass $m$.
For block $B$ not to slip down,the upward frictional force $f$ must balance the downward gravitational force $mg$.
So,$f = mg$.
The frictional force is given by $f = \mu N$,where $N$ is the normal force exerted by block $A$ on block $B$.
The normal force $N$ provides the acceleration $a$ to block $B$,so $N = ma$.
Substituting $a$: $N = m \left( \frac{F}{M+m} \right)$.
Now,equate the frictional force to the weight: $\mu N = mg$.
$\mu \left( m \frac{F}{M+m} \right) = mg$.
Solving for $F$: $F = \frac{(M+m)g}{\mu}$.
Thus,the correct option is $C$.

(A) Assume both blocks move together with a common acceleration $a$. The total mass of the system is $M = 1 \,kg + 100 \,kg = 101 \,kg$. The external force applied is $F = 10 \,N$.
Using Newton's second law,$F = Ma$,we get $10 = 101 \times a$,so $a = \frac{10}{101} \approx 0.099 \,m/s^2 \approx 0.1 \,m/s^2$.
The frictional force $f$ acting on the $1 \,kg$ block provides the acceleration to it. Thus,$f = m \times a = 1 \,kg \times 0.1 \,m/s^2 = 0.1 \,N$.
The maximum limiting friction is $f_L = \mu N = 0.5 \times 1 \,kg \times 10 \,m/s^2 = 5 \,N$.
Since the required frictional force $(0.1 \,N)$ is less than the limiting friction $(5 \,N)$,the blocks will move together,and the frictional force acting on the $1 \,kg$ block is $0.1 \,N$.

(D) $1$. Total mass of the system $(A + B) = 2 \, kg + 8 \, kg = 10 \, kg$.
$2$. The maximum static friction (limiting friction) between block $B$ and the horizontal floor is given by $f_{max} = \mu_s N$, where $N$ is the normal reaction force from the floor.
$3$. The normal reaction $N = (m_A + m_B)g = 10 \, kg \times 10 \, ms^{-2} = 100 \, N$.
$4$. Therefore, $f_{max} = 0.5 \times 100 \, N = 50 \, N$.
$5$. The applied horizontal force on block $B$ is $F_{app} = 10 \, N$.
$6$. Since $F_{app} < f_{max}$ $(10 \, N < 50 \, N)$, the entire system remains at rest.
$7$. Because the system is at rest and no external force is acting on block $A$ to cause relative motion between $A$ and $B$, the static friction force between blocks $A$ and $B$ is zero.

(A) The force required to make block $A$ slip over block $B$ is the limiting friction force $f_{max} = 12 \ N$.
For block $A$ to move with block $B$ without slipping,the maximum acceleration $a_{max}$ of the system is determined by the maximum friction force acting on $A$:
$f_{max} = m_A \cdot a_{max}$
$12 \ N = 4 \ kg \cdot a_{max}$
$a_{max} = 3 \ m \ s^{-2}$
Now,consider the system of both blocks $A$ and $B$ moving together with acceleration $a_{max}$. The total mass of the system is $M = m_A + m_B = 4 \ kg + 6 \ kg = 10 \ kg$.
The maximum horizontal force $F_B$ that can be applied on $B$ is:
$F_B = M \cdot a_{max}$
$F_B = 10 \ kg \cdot 3 \ m \ s^{-2} = 30 \ N$.

(D) $1$. Calculate the maximum static frictional force between block $A$ and slab $B$: $f_{s,max} = \mu_s N = \mu_s m_A g = 0.60 \times 10 \times 9.8 = 58 \ N$.
$2$. Since the applied force $(100 \ N)$ is greater than the maximum static friction $(58 \ N)$,block $A$ will slide over slab $B$.
$3$. Once sliding occurs,the kinetic friction force acts between the surfaces: $f_k = \mu_k N = \mu_k m_A g = 0.40 \times 10 \times 9.8 = 39.2 \ N$.
$4$. This kinetic friction force $f_k$ is the only horizontal force acting on slab $B$ (since the floor is frictionless).
$5$. Using Newton's second law for slab $B$: $F_{net} = M_B a_B \Rightarrow f_k = M_B a_B$.
$6$. $39.2 = 30 \times a_B \Rightarrow a_B = \frac{39.2}{30} \approx 1.31 \ m \ s^{-2}$.

(C) Given:
Mass of block $A$,$m_A = 100 \ kg$
Mass of block $B$,$m_B = 300 \ kg$
Coefficient of friction between $A$ and $B$,$\mu_1 = 0.35$
Coefficient of friction between $B$ and the surface,$\mu_2 = 0.5$
Acceleration due to gravity,$g = 9.8 \ m/s^2$
When block $B$ is pulled with force $P$,block $A$ remains stationary due to the string attached to the wall. Thus,kinetic friction acts at both interfaces.
$1$. Frictional force between $A$ and $B$ $(f_1)$:
$f_1 = \mu_1 N_1 = \mu_1 m_A g$
$f_1 = 0.35 \times 100 \times 9.8 = 343 \ N$
$2$. Frictional force between $B$ and the surface $(f_2)$:
The normal force on the surface is the weight of both blocks: $N_2 = (m_A + m_B)g$
$f_2 = \mu_2 N_2 = 0.5 \times (100 + 300) \times 9.8$
$f_2 = 0.5 \times 400 \times 9.8 = 1960 \ N$
$3$. Total force $P$ required to move block $B$:
$P = f_1 + f_2$
$P = 343 + 1960 = 2303 \ N$
Comparing with the given options,the nearest value is $2350 \ N$. Therefore,option $(C)$ is the correct choice.

(B) The static friction force between the $10 \,kg$ and $40 \,kg$ block is given by:
$F_s = \mu_s R = 0.6 \times (10 \,kg) \times (9.8 \,ms^{-2}) = 58.8 \,N$
Since the applied force $(F = 100 \,N)$ is greater than the maximum static friction force $(58.8 \,N)$,the $10 \,kg$ block will move relative to the $40 \,kg$ slab.
Due to this relative motion,the kinetic friction force acting on the $40 \,kg$ slab is:
$f_k = \mu_k R = 0.4 \times (10 \,kg) \times (9.8 \,ms^{-2}) = 39.2 \,N$
This kinetic friction force $f_k$ is the only horizontal force acting on the $40 \,kg$ slab.
Therefore,the acceleration $a$ of the $40 \,kg$ slab is:
$a = \frac{f_k}{m_{slab}} = \frac{39.2 \,N}{40 \,kg} = 0.98 \,ms^{-2}$

(D) Let the mass of the upper block be $m_1 = 2 \,kg$ and the lower block be $m_2 = 4 \,kg$. The coefficient of friction between the blocks is $\mu = 0.5$.
First,calculate the limiting friction force $f_L$ between the two blocks:
$f_L = \mu N = \mu m_1 g = 0.5 \times 2 \,kg \times 10 \,ms^{-2} = 10 \,N$.
Now,consider the forces acting on the $2 \,kg$ block. $A$ horizontal force of $2 \,N$ is applied to it. Since the applied force $(2 \,N)$ is less than the limiting friction $(10 \,N)$,the $2 \,kg$ block will not slide relative to the $4 \,kg$ block.
In this state of static equilibrium,the friction force $f$ acting on the $2 \,kg$ block must exactly balance the applied external force to keep it stationary relative to the $4 \,kg$ block.
Therefore,$f = 2 \,N$.

(C) Since the blocks move together with no relative sliding,the acceleration $a$ of the system is given by Newton's second law: $a = \frac{F}{M+m}$.
The friction force $f$ acting on the upper block of mass $m$ provides the necessary acceleration for it to move with the lower block. Thus,$f = ma = m \left( \frac{F}{M+m} \right) = \frac{mF}{M+m}$.
The distance $s$ covered by the blocks in time $t$ starting from rest is $s = \frac{1}{2} a t^2 = \frac{1}{2} \left( \frac{F}{M+m} \right) t^2$.
The work done by friction on the upper block is $W = f \times s = \left( \frac{mF}{M+m} \right) \times \left( \frac{1}{2} \frac{F t^2}{M+m} \right) = \frac{m F^2 t^2}{2(M+m)^2}$.
Note: The work done by friction on the lower block is equal in magnitude but opposite in sign $(-W)$,so the net work done by friction on the system is zero. The question asks for the work done by friction on the blocks (implying the upper block),which is $\frac{m F^2 t^2}{2(M+m)^2}$.

(B) For both blocks to move together,the block $m_2$ must move with the same acceleration $a$ as $m_1$. The only force causing $m_2$ to accelerate is the static friction force $f$ acting between $m_1$ and $m_2$.
The maximum static friction force is $f_{\max} = \mu N = \mu m_2 g$.
Applying Newton's second law to block $m_2$:
$f_{\max} = m_2 a \implies \mu m_2 g = m_2 a \implies a = \mu g$.
Now,applying Newton's second law to the system of both blocks $(m_1 + m_2)$:
$F_{\max} = (m_1 + m_2) a$.
Substituting the value of $a$:
$F_{\max} = (m_1 + m_2) \mu g = \mu(m_1 + m_2) g$.

(C) Total mass of the system $M = 8 \text{ ton} + 2 \text{ ton} = 10 \text{ ton} = 10000 \text{ kg}$.
Breaking force $F = 25000 \text{ N}$.
Deceleration of the truck $a = F / M = 25000 / 10000 = 2.5 \text{ m/s}^2$.
The block of mass $m = 2000 \text{ kg}$ experiences a pseudo force $F_p = m \times a = 2000 \times 2.5 = 5000 \text{ N}$ in the forward direction.
The maximum static frictional force available is $f_{max} = \mu \times m \times g = 0.3 \times 2000 \times 10 = 6000 \text{ N}$.
Since the required force to keep the block stationary $(5000 \text{ N})$ is less than the maximum static friction $(6000 \text{ N})$, the block will not slide.
Therefore, the frictional force acting on the block is equal to the pseudo force, which is $5000 \text{ N}$.

(A) Let $m_1 = 0.8 \text{ kg}$ and $m_2 = 0.2 \text{ kg}$. The acceleration is $a = 5 \text{ m/s}^2$. The coefficient of friction is $\mu = 0.1$.
For the $0.2 \text{ kg}$ block, the tension $T$ pulls it towards the pulley, and the friction force $f_2$ from the $0.8 \text{ kg}$ block opposes this motion. The normal force on the $0.2 \text{ kg}$ block is $N_2 = m_2 g = 0.2 \times 10 = 2 \text{ N}$.
Equation of motion for $m_2$: $T - \mu N_2 = m_2 a \implies T - 0.1 \times 2 = 0.2 \times 5 \implies T - 0.2 = 1.0 \implies T = 1.2 \text{ N}$.
For the $0.8 \text{ kg}$ block, the applied force $F$ acts to the right. The tension $T$ acts to the left. The friction force $f_1$ from the table acts to the left, and the friction force $f_2$ from the $0.2 \text{ kg}$ block also acts to the left. The normal force from the table is $N_1 = (m_1 + m_2)g = (0.8 + 0.2) \times 10 = 10 \text{ N}$.
Equation of motion for $m_1$: $F - T - \mu N_1 - \mu N_2 = m_1 a \implies F - 1.2 - 0.1 \times 10 - 0.1 \times 2 = 0.8 \times 5 \implies F - 1.2 - 1 - 0.2 = 4 \implies F - 2.4 = 4 \implies F = 6.4 \text{ N}$.

(A) Let $m_A = 3 \ kg$ and $m_B = 7 \ kg$. The applied force $F = 50 \ N$ acts on block $B$.
First,we check if the system moves. The maximum static friction force between block $B$ and the floor is $f_{max, floor} = \mu_{floor} (m_A + m_B) g$.
$f_{max, floor} = 0.55 \times (3 + 7) \times 10 = 0.55 \times 100 = 55 \ N$.
Since the applied force $F = 50 \ N$ is less than the maximum static friction force $f_{max, floor} = 55 \ N$,the system remains at rest.
Because block $B$ does not move and there is no external force acting on block $A$ to cause it to slide relative to block $B$,the static friction force between $A$ and $B$ must be zero to maintain equilibrium.
Therefore,the force of friction between $A$ and $B$ is $0 \ N$.

(B) Let $N_{1}$ be the normal force between the two blocks and $N_{2}$ be the normal force between the lower block and the table.
For the upper block of mass $m_{1}$,the normal force is $N_{1} = m_{1}g$.
For the lower block of mass $m_{2}$,the total downward force is $N_{2} = m_{2}g + N_{1} = (m_{1} + m_{2})g$.
The horizontal force $F$ on the upper block creates a frictional force $f_{1}$ on the lower block,where $f_{1} \leq \mu_{1}N_{1} = \mu_{1}m_{1}g$.
For the lower block to never move,the maximum static frictional force from the table must be greater than or equal to the maximum possible frictional force exerted by the upper block on the lower block.
Thus,$\mu_{2}N_{2} \geq \mu_{1}N_{1}$.
Substituting the values,$\mu_{2}(m_{1} + m_{2})g \geq \mu_{1}m_{1}g$.
Dividing both sides by $\mu_{2}m_{1}g$,we get $\frac{m_{1} + m_{2}}{m_{1}} \geq \frac{\mu_{1}}{\mu_{2}}$.
Therefore,$\frac{\mu_{1}}{\mu_{2}} \leq 1 + \frac{m_{2}}{m_{1}}$.
The maximum possible value is $1 + \frac{m_{2}}{m_{1}}$.