Newton's Laws of Motion

Step 1: Check the friction force between block A ($5\text{ kg}$) and block B ($10\text{ kg}$). Since the coefficient of friction is $\mu = 0$, the maximum friction force between block A and block B is $f_{\text{AB}} = \mu N_{\text{A}} = 0 \times (5 \times 10) = 0\text{ N}$.
Step 2: Calculate the acceleration of block A ($m_{\text{A}} = 5\text{ kg}$) under the external horizontal force $F = 15\text{ N}$: $a_{\text{A}} = \frac{F}{m_{\text{A}}} = \frac{15}{5} = 3\text{ m/s}^2$.
Step 3: Analyze the horizontal forces on block B ($10\text{ kg}$). Since block A exerts zero friction on block B ($f_{\text{AB}} = 0$) and no other external force acts on block B, the net horizontal force on block B is zero.
Step 4: Determine the acceleration and ground friction for block B. Since net force on B is zero, block B remains at rest ($a_{\text{B}} = 0$) and the friction force between block B and Ground G is $f_{\text{BG}} = 0\text{ N}$.

Step 1: Calculate the maximum available static friction force between block A ($5\text{ kg}$) and block B ($10\text{ kg}$):
$f_{\text{AB, max}} = \mu m_{\text{A}} g = 0.5 \times 5 \times 10 = 25\text{ N}$.
Step 2: Calculate the maximum available static friction force between block B and Ground G:
$f_{\text{BG, max}} = \mu' (m_{\text{A}} + m_{\text{B}}) g = 0.5 \times (5 + 10) \times 10 = 75\text{ N}$.
Step 3: Analyze the force applied on block A ($F = 30\text{ N}$). Since $F > f_{\text{AB, max}}$ ($30\text{ N} > 25\text{ N}$), block A slips relative to block B, and the friction force acting between A and B reaches its maximum value:
$f_{\text{AB}} = 25\text{ N}$.
Step 4: Calculate the acceleration of block A:
$a_{\text{A}} = \frac{F - f_{\text{AB}}}{m_{\text{A}}} = \frac{30 - 25}{5} = 1\text{ m/s}^2$.
Step 5: Analyze the motion of block B. The force driving block B forward is $f_{\text{AB}} = 25\text{ N}$. Since this force is less than the maximum ground static friction ($25\text{ N} < 75\text{ N}$), block B cannot overcome ground friction and remains at rest:
$a_{\text{B}} = 0$.
Step 6: Determine the static friction force exerted by the ground on block B to balance the forward force $f_{\text{AB}}$:
$f_{\text{BG}} = f_{\text{AB}} = 25\text{ N}$.

Step 1: Calculate the maximum limiting friction between block A ($5\text{ kg}$) and block B ($10\text{ kg}$):
$f_{\text{AB, max}} = \mu m_{\text{A}} g = 0.5 \times 5 \times 10 = 25\text{ N}$.
Step 2: Calculate the maximum limiting friction between block B and Ground G:
$f_{\text{BG, max}} = \mu' (m_{\text{A}} + m_{\text{B}}) g = 0.5 \times (5 + 10) \times 10 = 75\text{ N}$.
Step 3: Determine the maximum possible acceleration of block A without slipping:
$a_{\text{A, max}} = \frac{f_{\text{AB, max}}}{m_{\text{A}}} = \frac{25}{5} = 5\text{ m/s}^2$.
Step 4: Check if the blocks can move together under applied force $F = 200\text{ N}$ on block B. If they move together, the common acceleration would be $a = \frac{F - f_{\text{BG, max}}}{m_{\text{A}} + m_{\text{B}}} = \frac{200 - 75}{15} = 8.33\text{ m/s}^2$. Since $8.33\text{ m/s}^2 > 5\text{ m/s}^2$, relative slipping occurs between block A and block B.
Step 5: Calculate individual accelerations and friction forces during slipping:
Friction force between A and B is at its maximum value: $f_{\text{AB}} = 25\text{ N}$.
Friction force between B and Ground is at its maximum value: $f_{\text{BG}} = 75\text{ N}$.
Acceleration of block A: $a_{\text{A}} = \frac{f_{\text{AB}}}{m_{\text{A}}} = \frac{25}{5} = 5\text{ m/s}^2$.
Acceleration of block B: $a_{\text{B}} = \frac{F - f_{\text{AB}} - f_{\text{BG}}}{m_{\text{B}}} = \frac{200 - 25 - 75}{10} = \frac{100}{10} = 10\text{ m/s}^2$.

Step 1: Calculate the maximum limiting friction between block A ($5\text{ kg}$) and block B ($10\text{ kg}$):
$f_{\text{AB, max}} = \mu m_{\text{A}} g = 0.5 \times 5 \times 10 = 25\text{ N}$.
Step 2: Calculate the maximum limiting friction between block B and Ground G:
$f_{\text{BG, max}} = \mu' (m_{\text{A}} + m_{\text{B}}) g = 0.5 \times (5 + 10) \times 10 = 75\text{ N}$.
Step 3: Assume both blocks move together as a single unit and calculate the common acceleration:
$a = \frac{F - f_{\text{BG, max}}}{m_{\text{A}} + m_{\text{B}}} = \frac{90 - 75}{5 + 10} = \frac{15}{15} = 1\text{ m/s}^2$.
Step 4: Calculate the required friction force between block A and block B to accelerate block A at $1\text{ m/s}^2$:
$f_{\text{AB, req}} = m_{\text{A}} \times a = 5 \times 1 = 5\text{ N}$.
Step 5: Compare $f_{\text{AB, req}}$ with $f_{\text{AB, max}}$. Since $5\text{ N} \le 25\text{ N}$, the available friction is sufficient to move both blocks together without relative slipping.
Step 6: State the final values:
$a_{\text{A}} = a_{\text{B}} = 1\text{ m/s}^2$
$f_{\text{AB}} = 5\text{ N}$
$f_{\text{BG}} = 75\text{ N}$

Step 1: Calculate the common acceleration for all cases, as total mass $m_{\text{total}} = 2 + 4 = 6\text{ kg}$ and net external force $F_{\text{net}} = 12\text{ N}$ towards right in all situations:
$a = \frac{12}{6} = 2\text{ m/s}^2$ (towards right)
Step 2: Determine the friction force $f$ acting on the $2\text{ kg}$ block for each situation:
* Situation (p): $f = m_1 \cdot a = 2 \times 2 = 4\text{ N}$ towards right.
* Situation (q): $12 - f = 2 \times 2 \implies f = 8\text{ N}$ towards left.
* Situation (r): $6 - f = 2 \times 2 \implies f = 2\text{ N}$ towards left.
* Situation (s): $f - 8 = 2 \times 2 \implies f = 12\text{ N}$ towards right.
Step 3: Match the statements in Column-I with Column-II based on the calculated friction forces:
* (A) Magnitude of frictional force is maximum ($12\text{ N}$) $\rightarrow$ (s)
* (B) Magnitude of friction force is least ($2\text{ N}$) $\rightarrow$ (r)
* (C) Friction force on $2\text{ kg}$ block is towards right $\rightarrow$ (p), (s)
* (D) Friction force on $2\text{ kg}$ block is towards left $\rightarrow$ (q), (r)