Newton's Laws of Motion

Step 1: Identify the tension force acting on each body in the system. The leftmost block experiences an upward tension force $T$ and moves upwards with acceleration $a'$.
Step 2: The second block is connected to a movable pulley (providing $2T$ upward force) and moves upwards with acceleration $a = 2\text{ m/s}^2$, giving a tension work contribution of $2T \times 2 = 4T$.
Step 3: The third block is connected to a movable pulley (providing $2T$ upward force) and moves downwards with acceleration $a = 6\text{ m/s}^2$, giving a tension work contribution of $-2T \times 6 = -12T$.
Step 4: The rightmost platform is connected to a movable pulley (providing $2T$ upward force) and directly to the rightmost string segment (providing $T$ upward force), totaling an upward force of $3T$. It moves downwards with acceleration $a = 4\text{ m/s}^2$, giving a tension work contribution of $-3T \times 4 = -12T$.
Step 5: Write the virtual work constraint equation $\sum \vec{T} \cdot \vec{a} = 0$: $T a' + 4T - 12T - 12T = 0$.
Step 6: Simplify the terms by dividing by $T$: $a' + 4 - 12 - 12 = 0$.
Step 7: Combine like terms: $a' - 20 = 0$.
Step 8: Solve for $a'$ to get $a' = 20\text{ m/s}^2$.

Step 1: Identify the forces acting on each block. The $4\text{ kg}$ block experiences upward tension $T$ and downward gravity force $40\text{ N}$. The $10\text{ kg}$ block experiences upward tension $2T$ and downward gravity force $100\text{ N}$.
Step 2: Apply the string constraint relation using virtual work $\sum \vec{T} \cdot \vec{a} = 0$: $T a_2 - 2T a_1 = 0$, which gives $a_2 = 2 a_1$.
Step 3: Write Newton's second law equation for the $4\text{ kg}$ block moving upwards with acceleration $a_2$: $T - 40 = 4 a_2$.
Step 4: Write Newton's second law equation for the $10\text{ kg}$ block moving downwards with acceleration $a_1$: $100 - 2T = 10 a_1$.
Step 5: Substitute $a_2 = 2 a_1$ into the first equation to express tension in terms of $a_1$: $T - 40 = 4(2 a_1)$, which gives $T = 40 + 8 a_1$.
Step 6: Substitute $T = 40 + 8 a_1$ into the second equation: $100 - 2(40 + 8 a_1) = 10 a_1$.
Step 7: Simplify the algebraic equation: $100 - 80 - 16 a_1 = 10 a_1$, which gives $20 = 26 a_1$.
Step 8: Solve for $a_1$: $a_1 = \frac{20}{26} = \frac{10}{13}\text{ m/s}^2$.
Step 9: Calculate $a_2$: $a_2 = 2 a_1 = 2 \times \frac{10}{13} = \frac{20}{13}\text{ m/s}^2$.

Step 1: Identify the forces acting on each block. The $4\text{ kg}$ block experiences an upward tension of $2T$ and a downward gravitational force of $40\text{ N}$, so its acceleration is $a_1 = \frac{2T - 40}{4}$.
Step 2: The $6\text{ kg}$ block experiences an upward tension of $T$ and a downward gravitational force of $60\text{ N}$, so its acceleration is $a_2 = \frac{T - 60}{6}$.
Step 3: Apply the virtual work principle for tension $\sum \vec{T} \cdot \vec{a} = 0$: $2T a_1 + T a_2 = 0$, which simplifies to $2a_1 + a_2 = 0$.
Step 4: Substitute the equations for $a_1$ and $a_2$ into the constraint relation: $2\left(\frac{2T - 40}{4}\right) + \frac{T - 60}{6} = 0$.
Step 5: Simplify the first term to get $\frac{2T - 40}{2} + \frac{T - 60}{6} = 0$.
Step 6: Multiply the entire equation by $6$ to clear denominators: $3(2T - 40) + (T - 60) = 0$.
Step 7: Expand and combine like terms: $6T - 120 + T - 60 = 0$, which gives $7T - 180 = 0$, yielding $T = \frac{180}{7}\text{ N}$.
Step 8: Substitute $T = \frac{180}{7}$ back into $a_1$: $a_1 = \frac{2\left(\frac{180}{7}\right) - 40}{4} = \frac{\frac{360 - 280}{7}}{4} = \frac{80}{28} = \frac{20}{7}\text{ m/s}^2$.
Step 9: Substitute $T = \frac{180}{7}$ back into $a_2$: $a_2 = \frac{\frac{180}{7} - 60}{6} = \frac{\frac{180 - 420}{7}}{6} = \frac{-240}{42} = -\frac{40}{7}\text{ m/s}^2$.

Step 1: Identify the forces acting on each block. The $4\text{ kg}$ block experiences upward tension $T$ and downward gravity force $40\text{ N}$, so its acceleration is $a_2 = \frac{T - 40}{4}$.
Step 2: The $8\text{ kg}$ block is connected to the central movable pulley (providing $2T$ upward force) and directly to the rightmost string (providing $T$ upward force), making a total upward tension of $3T$. With downward gravity force $80\text{ N}$, its acceleration is $a_1 = \frac{3T - 80}{8}$.
Step 3: Apply the virtual work principle for tension $\sum \vec{T} \cdot \vec{a} = 0$: $T a_2 + 3T a_1 = 0$, which simplifies to $a_2 + 3 a_1 = 0$.
Step 4: Substitute the expressions for $a_2$ and $a_1$ into the constraint equation: $\frac{T - 40}{4} + 3\left(\frac{3T - 80}{8}\right) = 0$.
Step 5: Multiply the entire equation by $8$ to clear the denominators: $2(T - 40) + 3(3T - 80) = 0$.
Step 6: Expand and combine like terms: $2T - 80 + 9T - 240 = 0$, which simplifies to $11T - 320 = 0$, giving $T = \frac{320}{11}\text{ N}$.
Step 7: Substitute $T = \frac{320}{11}$ into the equation for $a_1$: $a_1 = \frac{3\left(\frac{320}{11}\right) - 80}{8} = \frac{\frac{960 - 880}{11}}{8} = \frac{80}{88} = \frac{10}{11}\text{ m/s}^2$.
Step 8: Calculate $a_2$ using $a_2 = -3 a_1$: $a_2 = -3\left(\frac{10}{11}\right) = -\frac{30}{11}\text{ m/s}^2$.

Step 1: Identify the upward tension acting on each mass block based on the pulley string connections. The $2\text{ kg}$ block experiences $2T$, the $4\text{ kg}$ block experiences $4T$, the $8\text{ kg}$ block experiences $T$, and the $16\text{ kg}$ block experiences $T$.
Step 2: Write down the acceleration of each block using Newton's second law ($a = \frac{T_{\text{net}} - mg}{m}$):
$a_1 = \frac{2T - 20}{2}$
$a_2 = \frac{4T - 40}{4}$
$a_3 = \frac{T - 80}{8}$
$a_4 = \frac{T - 160}{16}$
Step 3: Apply the virtual work principle for string tension $\sum \vec{T} \cdot \vec{a} = 0$: $2T a_1 + 4T a_2 + T a_3 + T a_4 = 0$, which simplifies to $2a_1 + 4a_2 + a_3 + a_4 = 0$.
Step 4: Substitute the expressions for acceleration into the constraint relation: $2\left(\frac{2T - 20}{2}\right) + 4\left(\frac{4T - 40}{4}\right) + \left(\frac{T - 80}{8}\right) + \frac{T - 160}{16} = 0$.
Step 5: Simplify the terms: $(2T - 20) + (4T - 40) + \frac{T - 80}{8} + \frac{T - 160}{16} = 0$.
Step 6: Combine the integer terms to get $6T - 60 + \frac{T - 80}{8} + \frac{T - 160}{16} = 0$.
Step 7: Multiply the entire equation by $16$ to eliminate fractions: $16(6T - 60) + 2(T - 80) + (T - 160) = 0$.
Step 8: Expand and combine terms: $96T - 960 + 2T - 160 + T - 160 = 0$, giving $99T - 1280 = 0$, so $T = \frac{1280}{99}\text{ N}$.
Step 9: Calculate each acceleration:
$a_1 = \frac{2\left(\frac{1280}{99}\right) - 20}{2} = \frac{1280 - 990}{99} = \frac{290}{99}\text{ m/s}^2$
$a_2 = \frac{4\left(\frac{1280}{99}\right) - 40}{4} = \frac{1280 - 990}{99} = \frac{290}{99}\text{ m/s}^2$
$a_3 = \frac{\frac{1280}{99} - 80}{8} = \frac{1280 - 7920}{792} = -\frac{6640}{792} = -\frac{830}{99}\text{ m/s}^2$
$a_4 = \frac{\frac{1280}{99} - 160}{16} = \frac{1280 - 15840}{1584} = -\frac{14560}{1584} = -\frac{910}{99}\text{ m/s}^2$

Step 1: Determine the tension force acting on each mass block. The $1\text{ kg}$ block experiences upward tension $2T$, the $2\text{ kg}$ block experiences upward tension $T$, and the $3\text{ kg}$ block experiences upward tension $T$.
Step 2: Express the acceleration of each block using Newton's second law:
$a_1 = \frac{2T - 10}{1}$
$a_2 = \frac{T - 20}{2}$
$a_3 = \frac{T - 30}{3}$
Step 3: Apply the virtual work constraint for tension forces $\sum \vec{T} \cdot \vec{a} = 0$: $2T a_1 + T a_2 + T a_3 = 0$, which simplifies to $2 a_1 + a_2 + a_3 = 0$.
Step 4: Substitute the expressions for acceleration into the constraint relation: $2\left(\frac{2T - 10}{1}\right) + \frac{T - 20}{2} + \frac{T - 30}{3} = 0$.
Step 5: Multiply the entire equation by $6$ to clear all denominators: $12(2T - 10) + 3(T - 20) + 2(T - 30) = 0$.
Step 6: Expand the terms: $24T - 120 + 3T - 60 + 2T - 60 = 0$.
Step 7: Combine like terms: $29T - 240 = 0$, which gives $T = \frac{240}{29}\text{ N}$.
Step 8: Calculate $a_1$: $a_1 = \frac{2\left(\frac{240}{29}\right) - 10}{1} = \frac{480 - 290}{29} = \frac{190}{29}\text{ m/s}^2$.
Step 9: Calculate $a_2$: $a_2 = \frac{\frac{240}{29} - 20}{2} = \frac{\frac{240 - 580}{29}}{2} = -\frac{340}{58} = -\frac{170}{29}\text{ m/s}^2$.
Step 10: Calculate $a_3$: $a_3 = \frac{\frac{240}{29} - 30}{3} = \frac{\frac{240 - 870}{29}}{3} = -\frac{630}{87} = -\frac{210}{29}\text{ m/s}^2$.

Step 1: Identify the tension force acting on each mass block. Block $2m$ on the horizontal surface is pulled to the right by a tension of $2T$, and hanging mass $m$ experiences an upward tension of $T$.
Step 2: Express the horizontal acceleration of block $2m$: $a_2 = \frac{2T}{2m}$.
Step 3: Express the vertical acceleration of mass $m$: $a_1 = \frac{T - mg}{m}$.
Step 4: Apply the virtual work constraint for tension forces $\sum \vec{T} \cdot \vec{a} = 0$: $2T a_2 + T a_1 = 0$, which simplifies to $2 a_2 + a_1 = 0$.
Step 5: Substitute the expressions for $a_2$ and $a_1$ into the constraint equation: $2\left(\frac{2T}{2m}\right) + \left(\frac{T - mg}{m}\right) = 0$.
Step 6: Cancel $m$ from the equation and simplify: $2T + T - mg = 0$.
Step 7: Combine tension terms to get $3T - mg = 0$, which yields $T = \frac{mg}{3}$.
Step 8: Substitute $T = \frac{mg}{3}$ into the expression for $a_1$: $a_1 = \frac{\frac{mg}{3} - mg}{m} = \frac{-\frac{2mg}{3}}{m} = -\frac{2g}{3}$ (the negative sign indicates a downward acceleration).

Step 1: Identify the forces acting on block $m$ along the inclined plane. The upward force along the incline is $T$ and the downward force is $mg \sin 30^\circ$, giving $a_1 = \frac{T - mg \sin 30^\circ}{m}$.
Step 2: Identify the forces acting on hanging block $2m$. The upward tension force is $2T$ and the downward gravitational force is $2mg$, giving $a_2 = \frac{2T - 2mg}{2m}$.
Step 3: Apply the virtual work constraint relation for string tension $\sum \vec{T} \cdot \vec{a} = 0$: $T a_1 + 2T a_2 = 0$, which simplifies to $a_1 + 2a_2 = 0$.
Step 4: Substitute the expressions for $a_1$ and $a_2$ into the constraint relation: $\frac{T - mg \sin 30^\circ}{m} + 2\left(\frac{2T - 2mg}{2m}\right) = 0$.
Step 5: Cancel mass $m$ from the denominators: $(T - mg \sin 30^\circ) + (2T - 2mg) = 0$.
Step 6: Substitute $\sin 30^\circ = \frac{1}{2}$: $T - \frac{1}{2}mg + 2T - 2mg = 0$.
Step 7: Combine like terms: $3T - \frac{5}{2}mg = 0$, which gives tension $T = \frac{5}{6}mg$.
Step 8: Substitute $T = \frac{5}{6}mg$ into $a_1$: $a_1 = \frac{\frac{5}{6}mg - \frac{1}{2}mg}{m} = \left(\frac{5}{6} - \frac{3}{6}\right)g = \frac{2}{6}g = \frac{g}{3}$.
Step 9: Substitute $T = \frac{5}{6}mg$ into $a_2$: $a_2 = \frac{2\left(\frac{5}{6}mg\right) - 2mg}{2m} = \frac{\frac{5}{3}mg - 2mg}{2m} = \frac{-\frac{1}{3}mg}{2m} = -\frac{g}{6}$.

Step 1: Analyze the tension forces acting on each block. The string tension is $T$. Pulley $P_1$ connected to block $A$ has two string segments, so block $A$ (mass $2m$) is pulled rightwards by tension $2T$ and leftwards by force $F$.
Step 2: Write the equation of motion for block $A$ (taking leftward acceleration as $a_A$): $F - 2T = 2m a_A \implies a_A = \frac{F - 2T}{2m}$.
Step 3: Block $B$ (mass $4m$) is attached to one string end (tension $T$) and pulley $P_2$ (tension $2T$), giving a total leftward tension force of $3T$. Its equation of motion is $3T = 4m a_B \implies a_B = \frac{3T}{4m}$.
Step 4: Block $C$ (mass $m$) is attached to the string coming from $P_2$, experiencing a rightward tension force $T$. Its rightward acceleration is $a_C = \frac{T}{m}$.
Step 5: Relate the positions using string length: $L = 2(X_B - X_A) + (X_B - X_C) = 3X_B - 2X_A - X_C$. Differentiating twice yields the constraint relation: $3a_B - 2a_A - a_C = 0$, where $a_A$ and $a_B$ are leftward accelerations and $a_C$ is rightward acceleration.
Step 6: Substitute $a_A$, $a_B$, and $a_C$ into the constraint equation: $3\left(\frac{3T}{4m}\right) = 2\left(\frac{F - 2T}{2m}\right) - \frac{T}{m}$.
Step 7: Cancel mass $m$ and simplify: $\frac{9T}{4} = F - 2T - T \implies \frac{9T}{4} = F - 3T$.
Step 8: Combine tension terms to solve for $T$: $\frac{9T}{4} + \frac{12T}{4} = F \implies \frac{21T}{4} = F \implies T = \frac{4F}{21}$.
Step 9: Substitute $T = \frac{4F}{21}$ into the acceleration equation for block $B$: $a_B = \frac{3}{4m}\left(\frac{4F}{21}\right) = \frac{F}{7m}$.

Step 1: Assign tension ratios in the string starting from block $E$. Let tension pulling block $E$ upwards be $4T$.
Step 2: Determine tension forces pulling each block based on pulley system configuration:
Block $A$: Upward tension is $8T$, moving upwards at $8\text{ m/s}$.
Block $B$: Upward tension is $T$, moving downwards at $2\text{ m/s}$.
Block $C$: Upward tension is $2T$, moving upwards at $2\text{ m/s}$.
Block $D$: Upward tension is $8T$, moving downwards at $1\text{ m/s}$.
Block $E$: Upward tension is $4T$, velocity is $V_E$ upwards.
Block $F$: Horizontal tension is $16T$, moving rightwards at $4\text{ m/s}$.
Step 3: Write the virtual work equation $\sum \vec{T} \cdot \vec{V} = 0$:
$8T(8) - T(2) + 2T(2) - 8T(1) + 4T V_E - 16T(4) = 0$
Step 4: Cancel $T$ from all terms in the equation:
$64 - 2 + 4 - 8 + 4 V_E - 64 = 0$
Step 5: Simplify and solve for $4 V_E$:
$4 V_E - 62 = 0 \implies 4 V_E = 62$
Step 6: Calculate $V_E$:
$V_E = \frac{62}{4} = \frac{31}{2}\text{ ms}^{-1}$

Step 1: Calculate the normal force $N = mg = 10 kg \times 10 m/s^2 = 100 N$.
Step 2: Calculate the maximum static friction using the formula $(f_{s})_{max} = \mu_s N = 0.4 \times 100 N = 40 N$.
Step 3: Compare the applied force ($F = 10 N$) with the maximum static friction ($40 N$). Since $10 N \le 40 N$, the block does not move.
Step 4: Determine the acceleration and friction for the stationary block: acceleration $a = 0$ and static friction $f = F = 10 N$.

Step 1: Calculate the normal force $N = mg = 10 kg \times 10 m/s^2 = 100 N$.
Step 2: Calculate the maximum static friction $(f_{s})_{max} = \mu_s N = 0.4 \times 100 N = 40 N$.
Step 3: Compare the applied force ($F = 20 N$) with the maximum static friction ($40 N$). Since $20 N \le 40 N$, the block remains at rest.
Step 4: Determine the acceleration and friction: acceleration $a = 0$ and static friction $f = F = 20 N$.
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Step 1: Calculate the normal force $N = mg = 10 kg \times 10 m/s^2 = 100 N$.
Step 2: Calculate the maximum static friction $(f_{s})_{max} = \mu_s N = 0.4 \times 100 N = 40 N$.
Step 3: Compare the applied force ($F = 35 N$) with the maximum static friction ($40 N$). Since $35 N \le 40 N$, the block remains at rest.
Step 4: Determine the acceleration and friction: acceleration $a = 0$ and static friction $f = F = 35 N$.
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Step 1: Calculate the normal force $N = mg = 10 kg \times 10 m/s^2 = 100 N$.
Step 2: Calculate the maximum static friction $(f_{s})_{max} = \mu_s N = 0.4 \times 100 N = 40 N$.
Step 3: Compare the applied force ($F = 39.99 N$) with the maximum static friction ($40 N$). Since $39.99 N \le 40 N$, the block remains at rest.
Step 4: Determine the acceleration and friction: acceleration $a = 0$ and static friction $f = F = 39.99 N$.
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Step 1: Calculate the kinetic friction $f_k = \mu_k N = 0.3 \times 100 N = 30 N$.
Step 2: Substitute the applied force and kinetic friction into the acceleration formula $a = \frac{F - f_k}{m}$.
Step 3: Compute $a = \frac{40.01 - 30}{10}$.
Step 4: Obtain the final acceleration value $a = 1.001 m/s^2$.
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Step 1: Calculate the kinetic friction $f_k = \mu_k N = 0.3 \times 100 N = 30 N$.
Step 2: Substitute the applied force and kinetic friction into the acceleration formula $a = \frac{F - f_k}{m}$.
Step 3: Compute $a = \frac{50 - 30}{10}$.
Step 4: Obtain the final acceleration value $a = 2 m/s^2$.

Step 1: Calculate the normal force $N = 10g + 100 = 200 N$.
Step 2: Calculate the maximum static friction $(f_{s})_{max} = \mu N = 0.1 \times 200 N = 20 N$.
Step 3: Find the net applied force magnitude $\vert{}F_2 - F_1\vert{} = \vert{}5 - 0\vert{} = 5 N$.
Step 4: Compare the net force with $(f_{s})_{max}$; since $5 N \le 20 N$, the static friction balances the net force, resulting in $f = 5 N$ towards left.
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Step 1: Determine the maximum static friction $(f_{s})_{max} = 20 N$.
Step 2: Calculate the net applied force magnitude $\vert{}F_2 - F_1\vert{} = \vert{}15 - 0\vert{} = 15 N$.
Step 3: Compare with $(f_{s})_{max}$; since $15 N \le 20 N$, the block is stationary.
Step 4: Conclude that the static friction equals the net force: $f = 15 N$ towards left.
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Step 1: Identify $(f_{s})_{max} = 20 N$.
Step 2: Calculate the net applied force magnitude $\vert{}F_2 - F_1\vert{} = \vert{}25 - 0\vert{} = 25 N$.
Step 3: Compare with $(f_{s})_{max}$; since $25 N > 20 N$, the block moves.
Step 4: Determine the kinetic friction $f = f_k = 20 N$ acting towards the left.
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Step 1: Find $(f_{s})_{max} = 20 N$.
Step 2: Compute the net applied force magnitude $\vert{}F_2 - F_1\vert{} = \vert{}10 - 20\vert{} = 10 N$.
Step 3: Check condition: $10 N \le 20 N$, so the block is stationary.
Step 4: Determine static friction $f = 10 N$ acting towards the right.
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Step 1: Calculate the net force magnitude $\vert{}F_2 - F_1\vert{} = \vert{}20 - 20\vert{} = 0$.
Step 2: Conclude that no friction is required to keep the system balanced, so $f = 0$.
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Step 1: Determine $(f_{s})_{max} = 20 N$.
Step 2: Compute the net applied force magnitude $\vert{}F_2 - F_1\vert{} = \vert{}42 - 20\vert{} = 22 N$.
Step 3: Since $22 N > 20 N$, the block moves.
Step 4: Find the kinetic friction $f = 20 N$ acting towards the left.

Step 1: Resolve the applied force into components where $F \cos 37^\circ = 10 \times \frac{4}{5} = 8 N$ and $F \sin 37^\circ = 10 \times \frac{3}{5} = 6 N$.
Step 2: Calculate the normal force $N = mg - F \sin 37^\circ = 100 - 6 = 94 N$.
Step 3: Calculate the maximum static friction $(f_{s})_{max} = \mu_s N = 0.3 \times 94 = 28.2 N$.
Step 4: Compare the horizontal force $8 N$ with $(f_{s})_{max} = 28.2 N$; since $8 N \le 28.2 N$, the block does not move, giving acceleration $a = 0$ and static friction $f = 8 N$.
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Step 1: Determine components for $F = 50 N$: $F \cos 37^\circ = 40 N$ and $F \sin 37^\circ = 30 N$.
Step 2: Calculate the normal force $N = 100 - 30 = 70 N$.
Step 3: Calculate the maximum static friction $(f_{s})_{max} = 0.3 \times 70 = 21 N$ and kinetic friction $f_k = 0.2 \times 70 = 14 N$.
Step 4: Check motion condition: $40 N > 21 N$, so the block accelerates.
Step 5: Compute acceleration $a = \frac{40 - 14}{10} = 2.6 m/s^2$ with kinetic friction $f = 14 N$.

Step 1: Resolve the applied force into components where horizontal force is $F \cos 37^\circ = 100 \times \frac{4}{5} = 80 N$ and vertical component is $F \sin 37^\circ = 100 \times \frac{3}{5} = 60 N$.
Step 2: Calculate the normal force $N = mg + F \sin 37^\circ = 100 + 60 = 160 N$.
Step 3: Calculate the maximum static friction $(f_{s})_{max} = \mu N = 0.2 \times 160 = 32 N$.
Step 4: Compute the acceleration $a = \frac{F \cos 37^\circ - (f_{s})_{max}}{m} = \frac{80 - 32}{10} = 4.8 m/s^2$.

Step 1: Resolve the applied force into components where horizontal force is $F \cos 37^\circ = 200 \times \frac{4}{5} = 160 N$ and vertical component is $F \sin 37^\circ = 200 \times \frac{3}{5} = 120 N$.
Step 2: Calculate the normal force $N = mg + F \sin 37^\circ = 100 + 120 = 220 N$.
Step 3: Calculate the maximum static friction $(f_{s})_{max} = \mu N = 0.2 \times 220 = 44 N$.
Step 4: Compute the acceleration $a = \frac{F \cos 37^\circ - (f_{s})_{max}}{m} = \frac{160 - 44}{10} = 11.6 m/s^2$.

Step 1: Calculate the maximum static friction $(f_{s})_{max} = \mu_s N = 0.4 \times 100 = 40 N$ and kinetic friction $f_k = \mu_k N = 0.3 \times 100 = 30 N$.
Step 2: Compare the pulling force from the hanging mass ($m_2 g = 5 \times 10 = 50 N$) with $(f_{s})_{max}$. Since $50 N > 40 N$, the system moves.
Step 3: Apply Newton's second law for the combined system to find acceleration: $a = \frac{m_2 g - f_k}{m_1 + m_2} = \frac{50 - 30}{10 + 5} = \frac{4}{3} m/s^2$.
Step 4: Calculate the tension in the string using the vertical motion of the hanging mass: $T = 50 - 5 \times \frac{4}{3} = \frac{130}{3} N$.

Step 1: Identify the given values from the problem statement: Mass of the block $m = 1\text{ kg}$, coefficient of friction $\mu = 0.2$, and acceleration due to gravity $g = 9.8\text{ m/s}^2$.
Step 2: Relate the perpendicular force $F$ to the normal force $N$ acting on the wall, where $N = F$.
Step 3: Set the maximum frictional force equal to the weight of the block to find the limiting equilibrium state: $f = \mu N = \mu F = mg$.
Step 4: Rearrange the formula to solve for the minimum force: $F = \frac{mg}{\mu}$.
Step 5: Substitute the values into the equation: $F = \frac{1 \times 9.8}{0.2} = 49\text{ N}$.

Step 1: Identify the given values: Mass of the block $m = 20\text{ kg}$, horizontal force required to start motion $F_s = 75\text{ N}$, and acceleration due to gravity $g = 9.8\text{ m/s}^2$.
Step 2: Calculate the normal force acting on the block using the weight formula: $N = mg = 20 \times 9.8 = 196\text{ N}$.
Step 3: Apply the coefficient of static friction formula: $\mu_s = \frac{F_s}{N}$.
Step 4: Substitute the values into the formula: $\mu_s = \frac{75}{196} \approx 0.383$.
Step 5: Round the calculated value to match the options provided: $\mu_s = 0.38$.

Step 1: Write down the given physical parameters: mass $m = 2\text{ kg}$, coefficient of static friction $\mu_s = 0.4$, applied force $F = 2.5\text{ N}$, and acceleration due to gravity $g = 9.8\text{ m/s}^2$.
Step 2: Calculate the normal reaction force: $N = mg = 2 \times 9.8 = 19.6\text{ N}$.
Step 3: Calculate the maximum limiting static friction force: $f_{max} = \mu_s N = 0.4 \times 19.6 = 7.84\text{ N}$.
Step 4: Compare the applied horizontal force $F$ with $f_{max}$: since $2.5\text{ N} < 7.84\text{ N}$, the force is insufficient to overcome static friction and move the block.
Step 5: Determine the actual frictional force: because static friction self-adjusts to balance the applied force when $F \le f_{max}$, the frictional force is equal to the applied force, $f = 2.5\text{ N}$.

Step 1: Identify the given values from the problem statement: Weight of the ladder $W = 250\text{ N}$ and coefficient of friction between the ladder and floor $\mu = 0.3$.
Step 2: Determine the normal reaction force $N$ exerted by the floor by resolving forces vertically: $N = W = 250\text{ N}$.
Step 3: Apply the maximum static friction formula: $f_{max} = \mu N$.
Step 4: Substitute the given values into the formula: $f_{max} = 0.3 \times 250\text{ N}$.
Step 5: Calculate the final result: $f_{max} = 75\text{ N}$.

Step 1: Identify the given values from the problem: mass $m = 2\text{ kg}$, pressing force $F = 100\text{ N}$, coefficient of friction $\mu = 0.3$, and $g \approx 10\text{ m/s}^2$ (or $9.8\text{ m/s}^2$).
Step 2: Calculate the normal reaction force $N$ supplied by the wall, which is equal to the pressing force: $N = 100\text{ N}$.
Step 3: Calculate the maximum limiting static friction force: $f_{max} = \mu N = 0.3 \times 100 = 30\text{ N}$.
Step 4: Calculate the weight of the body that acts downward: $W = mg = 2 \times 10 = 20\text{ N}$.
Step 5: Compare the weight $W$ with $f_{max}$: since $20\text{ N} \le 30\text{ N}$, the body does not slide, and the actual frictional force supporting the body equals its weight, $f = 20\text{ N}$.

Step 1: Identify the given values: Mass $m = 5\text{ kg}$, applied force $F = 24\text{ N}$, coefficient of kinetic friction $\mu_k = 0.4$, and acceleration due to gravity $g = 9.8\text{ m/s}^2$.
Step 2: Calculate the normal reaction force acting on the block: $N = mg = 5 \times 9.8 = 49\text{ N}$.
Step 3: Calculate the kinetic frictional force opposing the motion: $f_k = \mu_k N = 0.4 \times 49 = 19.6\text{ N}$.
Step 4: Find the net effective force acting on the block: $F_{net} = F - f_k = 24 - 19.6 = 4.4\text{ N}$.
Step 5: Apply Newton's second law of motion to calculate the acceleration: $a = \frac{F_{net}}{m} = \frac{4.4}{5} = 0.88\text{ m/s}^2$.

Step 1: Identify the given quantities: mass $m = 2\text{ kg}$, velocity $v = 2\text{ m/s}$, coefficient of friction $\mu = 0.20$, time interval $t = 5\text{ s}$, acceleration due to gravity $g = 9.8\text{ m/s}^2$, and mechanical equivalent of heat $J = 4.2\text{ J/cal}$.
Step 2: Calculate the frictional force acting on the body: $f = \mu mg = 0.20 \times 2 \times 9.8 = 3.92\text{ N}$.
Step 3: Calculate the total distance covered in 5 seconds at a constant speed: $s = v \times t = 2 \times 5 = 10\text{ m}$.
Step 4: Compute the work done against friction: $W = f \times s = 3.92 \times 10 = 39.2\text{ Joules}$.
Step 5: Convert the work done into heat energy in calories by dividing by $J$: $H = \frac{39.2}{4.2} \approx 9.33\text{ cal}$.

Step 1: Identify the given values from the problem and diagram: Mass $M = 5\text{ kg}$, coefficient of friction $\mu = 0.2$, applied force $F = 40\text{ N}$ at an angle $\theta = 30^\circ$ with the horizontal, and acceleration due to gravity $g = 10\text{ m/s}^2$.
Step 2: Resolve the applied force into horizontal component $F \cos 30^\circ = 40 \times \cos 30^\circ = 40 \times 0.866 = 34.64\text{ N}$ and vertical component $F \sin 30^\circ = 40 \times 0.5 = 20\text{ N}$.
Step 3: Calculate the normal reaction force $N$ by balancing vertical forces: $N + F \sin 30^\circ = Mg \implies N = 5 \times 10 - 20 = 30\text{ N}$.
Step 4: Compute the frictional force opposing motion: $f_k = \mu N = 0.2 \times 30 = 6\text{ N}$.
Step 5: Calculate the net horizontal force: $F_{net} = F \cos 30^\circ - f_k = 34.64 - 6 = 28.64\text{ N}$.
Step 6: Compute the acceleration of the block using Newton's second law: $a = \frac{F_{net}}{M} = \frac{28.64}{5} = 5.728\text{ m/sec}^2 \approx 5.73\text{ m/sec}^2$.

Step 1: Consider a body of mass $m$ sliding down a rough inclined plane of inclination angle $\theta$.
Step 2: Resolve the gravitational force $mg$ into two components: $mg \sin\theta$ parallel to the incline acting downwards, and $mg \cos\theta$ perpendicular to the incline.
Step 3: Determine the normal reaction force $N$ perpendicular to the inclined plane: $N = mg \cos\theta$.
Step 4: Calculate the frictional force $f$ acting upwards along the inclined surface: $f = \mu N = \mu mg \cos\theta$, where $\mu$ is the coefficient of kinetic friction.
Step 5: Formulate the net force equation along the incline: $F_{net} = mg \sin\theta - \mu mg \cos\theta = mg(\sin\theta - \mu \cos\theta)$.
Step 6: Apply Newton's second law ($F_{net} = ma$) to solve for the acceleration: $a = g(\sin\theta - \mu \cos\theta)$.

Step 1: Identify the given values from the problem statement: Applied horizontal force $F = 10\text{ N}$ and coefficient of friction $\mu = 0.2$.
Step 2: Determine the normal reaction force $N$ perpendicular to the vertical wall, which is equal to the horizontal pressing force: $N = F = 10\text{ N}$.
Step 3: State the equilibrium condition for vertical forces where the upward static friction force balances the downward weight of the block: $W = f = \mu N$.
Step 4: Substitute the values into the formula: $W = 0.2 \times 10\text{ N}$.
Step 5: Calculate the weight of the block: $W = 2\text{ N}$.

Step 1: Identify the forces acting on the connected system of three blocks ($m_1 = 10 \text{ kg}$, $m_2 = 10 \text{ kg}$, $m_3 = 20 \text{ kg}$).
Step 2: Calculate the pulling forces due to gravity on the hanging blocks, $F_1 = 10 \times 10 = 100 \text{ N}$ downwards on the left, and $F_3 = 20 \times 10 = 200 \text{ N}$ downwards on the right.
Step 3: Determine the kinetic friction force on the middle block resting on the horizontal surface, $f_k = \mu_k \cdot m_2 \cdot g = 0.4 \times 10 \times 10 = 40 \text{ N}$.
Step 4: Calculate the net pulling force causing motion towards the right, $F_{net} = 200 - 100 - 40 = 60 \text{ N}$.
Step 5: Calculate the total mass of the system, $m_{total} = 10 + 10 + 20 = 40 \text{ kg}$.
Step 6: Compute the common acceleration $a = \frac{F_{net}}{m_{total}} = \frac{60}{40} = \frac{3}{2} = 1.5 \text{ ms}^{-2}$.

Step 1: Determine the forces acting on the $10 \text{ kg}$ block on the $37^\circ$ incline.
Step 2: Calculate the normal force $N = mg \cos 37^\circ = 10 \times 10 \times 0.8 = 80 \text{ N}$.
Step 3: Calculate the downward weight component along the incline $mg \sin 37^\circ = 10 \times 10 \times 0.6 = 60 \text{ N}$.
Step 4: Determine maximum static friction force $(f_s)_{max} = \mu_s N = 0.6 \times 80 = 48 \text{ N}$ and kinetic friction force $f_k = \mu_k N = 0.5 \times 80 = 40 \text{ N}$.
Step 5: Identify the downward pulling force from the hanging mass $m_2 g = 110 \text{ N}$ (or $120 \text{ N}$ based on standard diagram parameters used in the step calculation).
Step 6: Determine net force causing motion $F_{net} = 110 - 60 - 40 = 10 \text{ N}$.
Step 7: Calculate total mass $m_{total} = 10 + 11 = 21 \text{ kg}$.
Step 8: Calculate acceleration $a = \frac{F_{net}}{m_{total}} = \frac{10}{21} \text{ ms}^{-2}$ and kinetic friction force $f = 40 \text{ N}$.

Step 1: Identify the forces acting on the block resting on the $37^\circ$ inclined plane ($m_1 = 10 \text{ kg}$).
Step 2: Calculate the normal reaction force $N = m_1 g \cos 37^\circ = 10 \times 10 \times 0.8 = 80 \text{ N}$.
Step 3: Calculate the downward weight component along the inclined plane $m_1 g \sin 37^\circ = 10 \times 10 \times 0.6 = 60 \text{ N}$.
Step 4: Compute the maximum static friction $(f_s)_{max} = \mu_s N = 0.6 \times 80 = 48 \text{ N}$ and kinetic friction $f_k = \mu_k N = 0.5 \times 80 = 40 \text{ N}$.
Step 5: Identify the downward gravitational force acting on the hanging mass ($m_2 = 11 \text{ kg}$), which is $m_2 g = 110 \text{ N}$.
Step 6: Determine the net pulling force causing motion upwards along the incline $F_{net} = 110 - 60 - 40 = 10 \text{ N}$.
Step 7: Find the total mass of the connected system $m_{total} = m_1 + m_2 = 10 + 11 = 21 \text{ kg}$.
Step 8: Compute the system acceleration $a = \frac{F_{net}}{m_{total}} = \frac{10}{21} \text{ ms}^{-2}$.

Step 1: Identify the normal force acting on the block: $N_1 = 100 \text{ N}$.
Step 2: Calculate the downward force due to gravity: $mg = 5 \times 10 = 50 \text{ N}$.
Step 3: Calculate the maximum static friction force: $(f_s)_{max} = \mu_s N_1 = 0.4 \times 100 = 40 \text{ N}$.
Step 4: Calculate the kinetic friction force: $f_k = \mu_k N_1 = 0.3 \times 100 = 30 \text{ N}$.
Step 5: Compare the downward force with maximum static friction: Since $mg = 50 \text{ N} > (f_s)_{max} = 40 \text{ N}$, the block slides downward.
Step 6: Compute acceleration: $a = \frac{mg - f_k}{m} = \frac{50 - 30}{5} = 4 \text{ m/s}^2$.
Step 7: Conclude that $f = f_k = 30 \text{ N}$ and acceleration $a = 4 \text{ m/s}^2$ downwards.
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Step 1: Identify the normal force acting on the block: $N_2 = 500 \text{ N}$.
Step 2: Calculate the downward force due to gravity: $mg = 5 \times 10 = 50 \text{ N}$.
Step 3: Calculate the maximum static friction force: $(f_s)_{max} = \mu_s N_2 = 0.4 \times 500 = 200 \text{ N}$.
Step 4: Compare downward force with maximum static friction: Since $mg = 50 \text{ N} \le (f_s)_{max} = 200 \text{ N}$, the block will not move.
Step 5: Determine static friction force balancing weight: $f = mg = 50 \text{ N}$.
Step 6: Conclude that $f = 50 \text{ N}$ and acceleration $a = 0 \text{ m/s}^2$.
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Step 1: Resolve the applied force $F = 500 \text{ N}$ at $37^\circ$ into horizontal and vertical components.
Step 2: Calculate normal force: $N_3 = F \cos 37^\circ = 500 \times 0.8 = 400 \text{ N}$.
Step 3: Calculate upward pushing force component: $F \sin 37^\circ = 500 \times 0.6 = 300 \text{ N}$.
Step 4: Calculate downward gravitational force: $mg = 12 \times 10 = 120 \text{ N}$.
Step 5: Calculate net upward driving force: $F_{driving} = 300 - 120 = 180 \text{ N}$.
Step 6: Calculate maximum static friction force: $(f_s)_{max} = \mu_s N_3 = 0.4 \times 400 = 160 \text{ N}$.
Step 7: Calculate kinetic friction force: $f_k = \mu_k N_3 = 0.3 \times 400 = 120 \text{ N}$.
Step 8: Compare driving force with static friction: Since $F_{driving} = 180 \text{ N} > (f_s)_{max} = 160 \text{ N}$, the block moves upwards.
Step 9: Compute acceleration upwards: $a = \frac{F_{driving} - f_k}{m} = \frac{180 - 120}{12} = 5 \text{ m/s}^2$.
Step 10: Conclude that $f = f_k = 120 \text{ N}$ and acceleration $a = 5 \text{ m/s}^2$ upwards.

Step 1: Identify the given values from the figure: mass $m = 10\text{ kg}$, coefficient of friction $\mu = 0.4$, initial velocity $u = 20\text{ m/s}$, and final velocity $V_f = 0\text{ m/s}$.
Step 2: Calculate the retardation acting on the block: $a = \mu g = 0.4 \times 10 = 4\text{ m/s}^2$ directed backwards.
Step 3: Apply the third equation of motion $V_f^2 = u^2 - 2ax$.
Step 4: Substitute the values into the equation: $0^2 = (20)^2 - 2 \times 4 \times x$.
Step 5: Solve for distance $x$: $0 = 400 - 8x \implies 8x = 400 \implies x = 50\text{ m}$.

Step 1: Identify the given parameters: initial speed $u = 50\text{ m/s}$, coefficient of friction $\mu = 0.5$, inclination angle $\theta = 37^\circ$, and final velocity $v = 0\text{ m/s}$.
Step 2: Calculate the retardation acting down the incline: $a = g\sin 37^\circ + \mu g\cos 37^\circ$.
Step 3: Substitute $g = 10\text{ m/s}^2$, $\sin 37^\circ = 0.6$, and $\cos 37^\circ = 0.8$: $a = 10(0.6) + 0.5 \times 10(0.8) = 6 + 4 = 10\text{ m/s}^2$.
Step 4: Apply the kinematic formula $v^2 = u^2 - 2ax$.
Step 5: Substitute the values into the formula: $0^2 = (50)^2 - 2 \times 10 \times x \implies 0 = 2500 - 20x$.
Step 6: Solve for distance $x$: $x = \frac{2500}{20} = 125\text{ m}$.

Step 1: Identify the forces acting on the hanging mass $m$: $T = mg$.
Step 2: Resolve the tension $T$ acting on the $10\text{ kg}$ block into horizontal component $T\cos 37^\circ$ and vertical component $T\sin 37^\circ$.
Step 3: Write the vertical equilibrium equation for the $10\text{ kg}$ block: $N + T\sin 37^\circ = 10 \times 10 = 100\text{ N} \implies N = 100 - T\sin 37^\circ$.
Step 4: Express maximum static friction force: $f_{max} = \mu N = 0.1 \times (100 - T\sin 37^\circ)$.
Step 5: Set horizontal pulling component equal to maximum static friction force for impending motion: $T\cos 37^\circ = 0.1(100 - T\sin 37^\circ)$.
Step 6: Substitute $\cos 37^\circ = 0.8$, $\sin 37^\circ = 0.6$, and $T = 10m$: $10m(0.8) = 0.1(100 - 10m(0.6))$.
Step 7: Simplify the algebraic equation: $8m = 10 - 0.6m \implies 8.6m = 10$.
Step 8: Solve for mass $m$: $m = \frac{10}{8.6} \approx 1.16\text{ kg}$.

Step 1: Calculate the maximum available static friction force between block A and block B using $f_{\text{max}} = \mu N = \mu m_{\text{A}} g = 0.5 \times 10 \times 10 = 50\text{ N}$.
Step 2: Assume both blocks move together as a single system of total mass $m_{\text{total}} = 10 + 10 = 20\text{ kg}$.
Step 3: Calculate the common acceleration $a = \frac{F}{m_{\text{total}}} = \frac{50}{20} = 2.5\text{ m/s}^2$.
Step 4: Determine the required friction force on block B to give it an acceleration of $2.5\text{ m/s}^2$ using $f = m_{\text{B}} \times a = 10 \times 2.5 = 25\text{ N}$.
Step 5: Compare the required friction force ($25\text{ N}$) with the maximum available static friction ($50\text{ N}$). Since $25\text{ N} \le 50\text{ N}$, the assumption is correct and both blocks move together with an acceleration of $a_{\text{A}} = a_{\text{B}} = 2.5\text{ m/s}^2$.

Step 1: Calculate the maximum available static friction force between block A and block B: $f_{\text{max}} = \mu m_{\text{A}} g = 0.5 \times 10 \times 10 = 50\text{ N}$.
Step 2: Assume both blocks move together as a single system and find the common acceleration: $a = \frac{F}{m_{\text{A}} + m_{\text{B}}} = \frac{101}{10 + 10} = 5.05\text{ m/s}^2$.
Step 3: Calculate the required friction force to move block B with acceleration $a = 5.05\text{ m/s}^2$: $f_{\text{required}} = m_{\text{B}} \times a = 10 \times 5.05 = 50.5\text{ N}$.
Step 4: Compare $f_{\text{required}}$ with $f_{\text{max}}$. Since $50.5\text{ N} > 50\text{ N}$, the required friction exceeds the maximum available friction, so the blocks move separately and slipping occurs.
Step 5: Calculate individual accelerations under kinetic friction $f_{\text{k}} = 50\text{ N}$:
For Block A: $a_{\text{A}} = \frac{F - f_{\text{k}}}{m_{\text{A}}} = \frac{101 - 50}{10} = 5.1\text{ m/s}^2$.
For Block B: $a_{\text{B}} = \frac{f_{\text{k}}}{m_{\text{B}}} = \frac{50}{10} = 5\text{ m/s}^2$.

Step 1: Calculate the maximum available static friction force between the $10\text{ kg}$ block and the $20\text{ kg}$ block: $f_{\text{max}} = \mu N = 0.5 \times 10 \times 10 = 50\text{ N}$.
Step 2: Assume both blocks move together as a single system of total mass $m_{\text{total}} = 10 + 20 = 30\text{ kg}$.
Step 3: Calculate the common acceleration of the combined system: $a = \frac{F}{m_{\text{total}}} = \frac{60}{30} = 2\text{ m/s}^2$.
Step 4: Calculate the required static friction force needed to move the bottom $20\text{ kg}$ block at this acceleration: $f_{\text{required}} = m \times a = 20 \times 2 = 40\text{ N}$.
Step 5: Compare the required friction force ($40\text{ N}$) with the maximum available static friction ($50\text{ N}$). Since $40\text{ N} < 50\text{ N}$, the required friction is less than the available static friction, so both blocks move together with an acceleration of $2\text{ m/s}^2$.

Step 1: Calculate the maximum limiting friction force between the $10\text{ kg}$ block and the $20\text{ kg}$ block: $f_{\text{max}} = \mu m_{\text{top}} g = 0.5 \times 10 \times 10 = 50\text{ N}$.
Step 2: Determine the acceleration of the top block ($10\text{ kg}$) when maximum limiting friction acts on it: $f_{\text{max}} = m_{\text{top}} \cdot a \implies 50 = 10 \cdot a \implies a = 5\text{ m/s}^2$.
Step 3: Write the equation of motion for the bottom block ($20\text{ kg}$) at the verge of sliding: $F - f_{\text{max}} = m_{\text{bottom}} \cdot a \implies F - 50 = 20 \cdot a$.
Step 4: Substitute $a = 5\text{ m/s}^2$ into the equation: $F - 50 = 20 \times 5 \implies F - 50 = 100 \implies F = 150\text{ N}$.
Step 5: Thus, the minimum horizontal force required to cause sliding between the two blocks is $F_{\text{min}} = 150\text{ N}$.