Question 1: Palette 1/105

DPT MCQ

Newton's Second Law of Motion, which states that the net force acting on an object is equal to the product of its mass and acceleration, represented by the formula $F_{net} = m a$.

From the image given below, find acceleration of blocks and the normal force between them.

A

$a = \frac{10}{7} \, \text{m/s}^2, N = \frac{50}{7} \, \text{N}$

B

$a = \frac{20}{7} \, \text{m/s}^2, N = \frac{100}{7} \, \text{N}$

C

$a = \frac{20}{7} \, \text{m/s}^2, N = \frac{50}{7} \, \text{N}$

D

$a = \frac{10}{7} \, \text{m/s}^2, N = \frac{100}{7} \, \text{N}$

Question 2: Palette 2/105

DPT MCQ

Newton's Second Law of Motion combined with equations of motion for connected bodies, represented by the formula $F_{net} = m_{total} a$ and individual force balance equations $F_{ext} - N = m_1 a$ and $N - F_{right} = m_2 a$.

From the image given below, find acceleration of the blocks and the normal force between them.

A

$a = 4 \, \text{m/s}^2, N = 12 \, \text{N}$

B

$a = 2 \, \text{m/s}^2, N = 10 \, \text{N}$

C

$a = 3 \, \text{m/s}^2, N = 14 \, \text{N}$

D

$a = 5 \, \text{m/s}^2, N = 16 \, \text{N}$

Question 3: Palette 3/105

DPT MCQ

Newton's Second Law of Motion and equations of motion for multiple connected bodies in contact, given by the formula $a = \frac{F_{net}}{m_{total}}$ and $N - F_{opp} = m_{block} a$.

From the image given below, find acceleration of blocks and normal force between 3Kg and 5Kg blocks.

A

$a = 4 \, \text{m/s}^2, N = 40 \, \text{N}$

B

$a = 2 \, \text{m/s}^2, N = 35 \, \text{N}$

C

$a = 4 \, \text{m/s}^2, N = 50 \, \text{N}$

D

$a = 6 \, \text{m/s}^2, N = 60 \, \text{N}$

Question 4: Palette 4/105

DPT MCQ

Newton's Second Law of Motion applied to multiple connected bodies, where the system acceleration is $a = \frac{F_{net}}{m_{total}}$ and the equation of motion for a subsystem is used to find the internal contact force $N - F_{opp} = m_{sub} a$.

From the image given below, find the interaction force between the $3 \, \text{kg}$ and $4 \, \text{kg}$ blocks.

A

$15 \, \text{N}$

B

$29 \, \text{N}$

C

$35 \, \text{N}$

D

$45 \, \text{N}$

Question 5: Palette 5/105

DPT MCQ

Newton's Second Law of Motion for connected bodies, where acceleration of the system is given by $a = \frac{F_{net}}{m_{total}}$ and the contact force $N$ is found using the equation of motion for an individual block $F_{applied} - N = m a$.

From the image given below, find the contact force between the $10 \, \text{kg}$ and $20 \, \text{kg}$ blocks.

A

$10 \, \text{N}$

B

$0 \, \text{N}$

C

$30 \, \text{N}$

D

$60 \, \text{N}$

Question 6: Palette 6/105

DPT MCQ

Newton's Second Law of Motion for connected bodies, where acceleration is given by $a = \frac{F}{m_1 + m_2}$ and the tension in the string is calculated using $T = m_1 a$.

From the image given below, find acceleration of each block and tension in the string.

A

$a = 10 \, \text{m/s}^2, T = 30 \, \text{N}$

B

$a = 20 \, \text{m/s}^2, T = 40 \, \text{N}$

C

$a = 15 \, \text{m/s}^2, T = 25 \, \text{N}$

D

$a = 25 \, \text{m/s}^2, T = 50 \, \text{N}$

Question 7: Palette 7/105

DPT MCQ

Newton's Second Law of Motion for connected systems of bodies and strings, given by the formulas $a = \frac{F}{m_1 + m_2 + m_3}$, $T_1 = m_1 a$, and $T_2 - T_1 = m_2 a$.

From the image given below, find the acceleration of the system and the tensions $T_1$ and $T_2$ in the strings.

A

$a = 10 \, \text{m/s}^2, T_1 = 20 \, \text{N}, T_2 = 50 \, \text{N}$

B

$a = 5 \, \text{m/s}^2, T_1 = 10 \, \text{N}, T_2 = 25 \, \text{N}$

C

$a = 15 \, \text{m/s}^2, T_1 = 30 \, \text{N}, T_2 = 60 \, \text{N}$

D

$a = 20 \, \text{m/s}^2, T_1 = 40 \, \text{N}, T_2 = 80 \, \text{N}$

Question 8: Palette 8/105

DPT MCQ

Newton's Second Law of Motion for connected bodies with opposing external forces, given by $a = \frac{F_{right} - F_{left}}{m_1 + m_2 + m_3}$, $T_1 - F_{left} = m_1 a$, and $F_{right} - T_2 = m_3 a$.

From the image given below, find the tensions $T_1$ and $T_2$.

A

$T_1 = 36 \, \text{N}, T_2 = 60 \, \text{N}$

B

$T_1 = 20 \, \text{N}, T_2 = 40 \, \text{N}$

C

$T_1 = 40 \, \text{N}, T_2 = 50 \, \text{N}$

D

$T_1 = 30 \, \text{N}, T_2 = 70 \, \text{N}$

Question 9: Palette 9/105

DPT MCQ

Newton's Second Law of Motion states that the net external force acting on a system is equal to the product of the total mass of the system and its acceleration, given by $F = m_{\text{total}} a$. The contact force between blocks can be found by applying Newton's second law to individual blocks or sub-systems using free-body diagrams, where the acceleration $a = \frac{F}{m_1 + m_2 + m_3}$ and the contact force $N$ accelerates the remaining blocks ahead of it.

from the image given below Find the acceleration of the system and the contact force between (i) $2\text{ kg}$ and $3\text{ kg}$ blocks (ii) $3\text{ kg}$ and $5\text{ kg}$ blocks

A

$a = 1\text{ m/s}^2$, $N_{23} = 6\text{ N}$, $N_{35} = 5\text{ N}$

B

$a = 1\text{ m/s}^2$, $N_{23} = 8\text{ N}$, $N_{35} = 5\text{ N}$

C

$a = 2\text{ m/s}^2$, $N_{23} = 8\text{ N}$, $N_{35} = 5\text{ N}$

D

$a = 1\text{ m/s}^2$, $N_{23} = 8\text{ N}$, $N_{35} = 6\text{ N}$

Question 10: Palette 10/105

DPT MCQ

Newton's Second Law of Motion relates force, mass, and acceleration as $F = m_{\text{total}} a$, where $a_{\text{system}} = \frac{P}{5m}$. The contact force between any two blocks is equal to the product of the mass of all subsequent blocks being pushed and the system acceleration $a$.

from the image given below Calculate : (i) $a_{\text{system}}$ (ii) $F_{\text{DE}}$ (iii) $F_{\text{CD}}$ (iv) $F_{\text{BC}}$ (v) $F_{\text{AB}}$ corresponding to the following diagram.

A

$a_{\text{system}} = \frac{P}{5m}$, $F_{\text{DE}} = \frac{P}{5}$, $F_{\text{CD}} = \frac{2P}{5}$, $F_{\text{BC}} = \frac{3P}{5}$, $F_{\text{AB}} = \frac{4P}{5}$

B

$a_{\text{system}} = \frac{P}{m}$, $F_{\text{DE}} = \frac{P}{5}$, $F_{\text{CD}} = \frac{P}{3}$, $F_{\text{BC}} = \frac{P}{2}$, $F_{\text{AB}} = \frac{2P}{3}$

C

$a_{\text{system}} = \frac{P}{5m}$, $F_{\text{DE}} = \frac{P}{2}$, $F_{\text{CD}} = \frac{P}{3}$, $F_{\text{BC}} = \frac{P}{4}$, $F_{\text{AB}} = \frac{P}{5}$

D

$a_{\text{system}} = \frac{2P}{5m}$, $F_{\text{DE}} = \frac{P}{5}$, $F_{\text{CD}} = \frac{2P}{5}$, $F_{\text{BC}} = \frac{3P}{5}$, $F_{\text{AB}} = P$

Question 11: Palette 11/105

DPT MCQ

Newton's Second Law of Motion relates the net external force acting on a multi-block system to its mass and acceleration by the formula $F_{\text{net}} = m_{\text{total}} a$, where the acceleration of the system is given by $a = \frac{P}{m_1 + m_2}$. The tension $T$ in the string connecting the two masses can be found by applying Newton's second law to either individual block, resulting in the formula $T = m_2 a = \frac{m_2 P}{m_1 + m_2}$.

from the image given below Calculate T for the following diagram :

A

$T = \frac{m_1 P}{m_1 + m_2}$

B

$T = \frac{m_2 P}{m_1 + m_2}$

C

$T = \frac{(m_1 + m_2)P}{m_1}$

D

$T = \frac{P}{m_1 m_2}$

Question 12: Palette 12/105

DPT MCQ

Newton's Second Law of Motion ($F_{\text{net}} = m \cdot a$). To find system acceleration, consider both blocks as a single system: $a = \frac{F_{\text{net}}}{\text{Total Mass}} = \frac{F_{\text{right}} - F_{\text{left}}}{m_1 + m_2}$. To find tension $T$, draw the Free Body Diagram (FBD) for one block and apply $F_{\text{net}} = m \cdot a$.

Two blocks of masses $2\text{ kg}$ and $3\text{ kg}$ connected by a string are pulled by forces $40\text{ N}$ to the left and $100\text{ N}$ to the right on a smooth horizontal surface. Find the acceleration of the system and the tension in the string connecting the blocks.

A

Acceleration = $12\text{ m/s}^2$, Tension = $64\text{ N}$

B

Acceleration = $10\text{ m/s}^2$, Tension = $50\text{ N}$

C

Acceleration = $12\text{ m/s}^2$, Tension = $40\text{ N}$

D

Acceleration = $8\text{ m/s}^2$, Tension = $64\text{ N}$

Question 13: Palette 13/105

DPT MCQ

Newton's Second Law of Motion and Motion of Blocks in Contact. The acceleration $a$ of the system is given by $a = \frac{F}{m_1 + m_2}$ and the force exerted on the lighter mass is $F' = m_2 a$.

from the image given below Two blocks of mass $4\ kg$ and $6\ kg$ are placed in contact with each other on a frictionless horizontal surface. If we apply a push of $5\ N$ on the heavier mass, the force on the lighter mass will be

A

$5\ N$

B

$4\ N$

C

$2\ N$

D

None of the above

Question 14: Palette 14/105

DPT MCQ

Newton's Second Law of Motion and contact forces between blocks. The force exerted on the other mass is given by the formula $F_{contact} = \frac{m_{other} F}{m_1 + m_2}$

from the image given below In the above problem, if a push of $5\ N$ is applied on the lighter mass, the force exerted by the lighter mass on the heavier mass will be

A

$5\ N$

B

$4\ N$

C

$3\ N$

D

None of the above

Question 15: Palette 15/105

DPT MCQ

Motion of blocks in contact under an applied force. The contact force $f$ between two blocks of masses $m_1$ and $m_2$ when a force $F$ is applied on $m_1$ is given by $f = \frac{m_2 F}{m_1 + m_2}$.

from the image given below Two blocks are in contact on a frictionless table one has a mass $m$ and the other $2\ m$ as shown in figure. Force $F$ is applied on mass $2\ m$ then system moves towards right. Now the same force $F$ is applied on $m$. The ratio of force of contact between the two blocks will be in the two cases respectively.

A

$1 : 3$

B

$1 : 2$

C

$1 : 1$

D

$1 : 4$

Question 16: Palette 16/105

DPT MCQ

Newton's Second Law of Motion and tension in connected blocks. The tension $T_1$ pulling the masses behind it is given by the formula $T_1 = \frac{(m_2 + m_3)F}{m_1 + m_2 + m_3}$

from the image given below Three blocks of masses $2\ kg$, $3\ kg$ and $5\ kg$ are connected to each other with light string and are then placed on a frictionless surface as shown in the figure. The system is pulled by a force $F = 10\ N$, then tension $T_1 =$

A

$1\ N$

B

$5\ N$

C

$8\ N$

D

$10\ N$

Question 17: Palette 17/105

DPT MCQ

Equilibrium of connected masses and pulleys, where the upward forces balance the downward gravitational forces, using the formulas $T = m_{lower} g$ and $2T = mg$.

From the image given below, find the mass $m$ if all masses are in equilibrium.

A

$10 \, \text{kg}$

B

$15 \, \text{kg}$

C

$20 \, \text{kg}$

D

$30 \, \text{kg}$

Question 18: Palette 18/105

DPT MCQ

Equilibrium of interconnected pulley systems, where tension is transmitted through strings and balances the gravitational forces on masses, using the formulas $T = 100 \, \text{N}$, $2T = m_1 g$, and $4T = m_2 g$.

From the image given below, find the unknown masses $m_1$ and $m_2$ when all masses are in equilibrium.

A

$m_1 = 10 \, \text{kg}, m_2 = 20 \, \text{kg}$

B

$m_1 = 20 \, \text{kg}, m_2 = 40 \, \text{kg}$

C

$m_1 = 30 \, \text{kg}, m_2 = 60 \, \text{kg}$

D

$m_1 = 40 \, \text{kg}, m_2 = 80 \, \text{kg}$

Question 19: Palette 19/105

DPT MCQ

Static equilibrium of connected pulley systems and strings, where the tension $T$ is uniform along a continuous string and balances the gravitational forces, using formulas $T = 10 \, g$, $2T = m_1 g$, and $T = m_2 g$.

From the image given below, find $m_1$ and $m_2$ if all masses are in equilibrium.

A

$m_1 = 10 \, \text{kg}, m_2 = 20 \, \text{kg}$

B

$m_1 = 20 \, \text{kg}, m_2 = 10 \, \text{kg}$

C

$m_1 = 20 \, \text{kg}, m_2 = 20 \, \text{kg}$

D

$m_1 = 10 \, \text{kg}, m_2 = 10 \, \text{kg}$

Question 20: Palette 20/105

DPT MCQ

Equilibrium of connected pulleys and an inclined plane system, where tension is multiplied through movable pulleys and balances the gravitational component along the incline using the formula $4T = mg \sin(30^\circ)$.

From the image given below, find the mass $m$ on the inclined plane if all masses are in equilibrium.

A

$40 \, \text{kg}$

B

$60 \, \text{kg}$

C

$80 \, \text{kg}$

D

$100 \, \text{kg}$

Question 21: Palette 21/105

DPT MCQ

Static equilibrium of a compound pulley system where a continuous string supports a heavy load, using the equilibrium conditions $3T = 300 \, \text{N}$ and $T = mg$.

From the image given below, find the unknown mass $m$ when all masses are in equilibrium.

A

$5 \, \text{kg}$

B

$10 \, \text{kg}$

C

$15 \, \text{kg}$

D

$20 \, \text{kg}$

Question 22: Palette 22/105

DPT MCQ

Static equilibrium of a connected pulley system using the principle of uniform string tension where $T = 10 \, g$, $T = m_2 g$, and $2T = m_1 g$.

From the image given below, find $m_1$ and $m_2$ if all masses are in equilibrium.

A

$m_1 = 10 \, \text{kg}, m_2 = 20 \, \text{kg}$

B

$m_1 = 20 \, \text{kg}, m_2 = 10 \, \text{kg}$

C

$m_1 = 30 \, \text{kg}, m_2 = 15 \, \text{kg}$

D

$m_1 = 40 \, \text{kg}, m_2 = 20 \, \text{kg}$

Question 23: Palette 23/105

DPT MCQ

Static equilibrium of a connected pulley system where a continuous string supports a mass on one side and a movable pulley/mass combination on the other, using the equilibrium conditions $T = 150 \, \text{N}$ and $3T = mg$.

From the image given below, find the unknown mass $m$ if all masses are in equilibrium.

A

$15 \, \text{kg}$

B

$30 \, \text{kg}$

C

$45 \, \text{kg}$

D

$60 \, \text{kg}$

Question 24: Palette 24/105

DPT MCQ

Static equilibrium of a multiple pulley system where a continuous string supports a $10 \, \text{kg}$ block on one end and three movable pulleys attached to mass $m$, using the formulas $T = 100 \, \text{N}$ and $6T = mg$.

From the image given below, find the unknown mass $m$ if all masses are in equilibrium.

A

$30 \, \text{kg}$

B

$40 \, \text{kg}$

C

$50 \, \text{kg}$

D

$60 \, \text{kg}$

Question 25: Palette 25/105

DPT MCQ

Equilibrium of a multi-pulley system connected to a hanging mass and a base platform, using the formulas $T = 100 \, \text{N}$ and $7T = mg$.

From the image given below, find the unknown mass $m$ if all masses are in equilibrium.

A

$50 \, \text{kg}$

B

$60 \, \text{kg}$

C

$70 \, \text{kg}$

D

$80 \, \text{kg}$

Question 26: Palette 26/105

DPT MCQ

Newton second law of motion applied to connected bodies moving vertically over a fixed pulley, using the equations of motion $m_1 a = m_1 g - T$ and $m_2 a = T - m_2 g$ to derive acceleration $a = \frac{(m_1 - m_2)g}{m_1 + m_2}$ and tension $T = \frac{2 m_1 m_2}{m_1 + m_2} g$.

From the image given below, when two bodies of mass $m_1$ and $m_2$ are attached at the ends of a string passing over a pulley (neglecting the mass of the pulley), find the acceleration $a$ of the system and the tension $T$ in the string.

A

$a = \frac{(m_1 - m_2)g}{m_1 + m_2}, T = \frac{2 m_1 m_2}{m_1 + m_2} g$

B

$a = \frac{(m_1 + m_2)g}{m_1 - m_2}, T = \frac{m_1 m_2}{m_1 + m_2} g$

C

$a = \frac{(m_1 - m_2)}{(m_1 + m_2)} g, T = \frac{m_1 + m_2}{2 m_1 m_2} g$

D

$a = \frac{2(m_1 - m_2)g}{m_1 + m_2}, T = \frac{m_1 m_2}{2(m_1 + m_2)} g$

Question 27: Palette 27/105

DPT MCQ

Atwood machine formula for the acceleration of connected masses $m_1$ and $m_2$ over an ideal pulley: $a = \frac{(m_1 - m_2)g}{m_1 + m_2}$.

From the image given below, find the acceleration of each mass if the pulley is ideal, the string is massless, and there is no friction between the pulley and the rope.

A

$1 \, \text{m/s}^2$

B

$2 \, \text{m/s}^2$

C

$3 \, \text{m/s}^2$

D

$4 \, \text{m/s}^2$

Question 28: Palette 28/105

DPT MCQ

Newton second law of motion for an Atwood machine, using the formulas $a = \frac{(m_1 - m_2)g}{m_1 + m_2}$ and $m_1 g - T = m_1 a$ to find acceleration and tension.

From the image given below, find the acceleration $a$ of the system and the string tension $T$ for masses $6 \, \text{kg}$ and $4 \, \text{kg}$ connected over an ideal pulley.

A

$a = 1 \, \text{m/s}^2, T = 36 \, \text{N}$

B

$a = 2 \, \text{m/s}^2, T = 48 \, \text{N}$

C

$a = 3 \, \text{m/s}^2, T = 42 \, \text{N}$

D

$a = 4 \, \text{m/s}^2, T = 50 \, \text{N}$

Question 29: Palette 29/105

DPT MCQ

Newton second law of motion for a mass on a frictionless horizontal table connected to a hanging mass, using the acceleration formula $a = \frac{m_1 g}{m_1 + m_2}$ and the tension equation $T = m_2 a$.

From the image given below, find the acceleration $a$ of the system and the string tension $T$ when a $4 \, \text{kg}$ block on a smooth horizontal surface is connected by a string over a pulley to a hanging $6 \, \text{kg}$ block.

A

$a = 4 \, \text{m/s}^2, T = 24 \, \text{N}$

B

$a = 5 \, \text{m/s}^2, T = 30 \, \text{N}$

C

$a = 6 \, \text{m/s}^2, T = 36 \, \text{N}$

D

$a = 8 \, \text{m/s}^2, T = 40 \, \text{N}$

Question 30: Palette 30/105

DPT MCQ

Newton second law for a combined system of a hanging mass and a mass on an inclined plane, using the formulas $a = \frac{m_1 g - m_2 g \sin\theta}{m_1 + m_2}$ and $m_1 g - T = m_1 a$.

From the image given below, find the acceleration $a$ of the system and the string tension $T$ for a $10 \, \text{kg}$ block on an inclined plane of $30^\circ$ connected to a hanging $20 \, \text{kg}$ block.

A

$a = 3 \, \text{m/s}^2, T = 120 \, \text{N}$

B

$a = 4 \, \text{m/s}^2, T = 140 \, \text{N}$

C

$a = 5 \, \text{m/s}^2, T = 100 \, \text{N}$

D

$a = 6 \, \text{m/s}^2, T = 80 \, \text{N}$

Question 31: Palette 31/105

DPT MCQ

Atwood machine dynamics with two fixed pulleys and a spring balance, using the acceleration formula $a = \frac{m_1 g - m_2 g}{m_1 + m_2}$ and the equation of motion $m_1 g - T = m_1 a$ to determine the tension $T$.

From the image given below, find the acceleration $a$ of the system and the spring balance reading $T$ when a $3 \, \text{kg}$ block and a $2 \, \text{kg}$ block are connected over two fixed pulleys with a spring balance in between.

A

$a = 1 \, \text{m/s}^2, T = 20 \, \text{N}$

B

$a = 2 \, \text{m/s}^2, T = 24 \, \text{N}$

C

$a = 3 \, \text{m/s}^2, T = 18 \, \text{N}$

D

$a = 4 \, \text{m/s}^2, T = 22 \, \text{N}$

Question 32: Palette 32/105

DPT MCQ

Newton second law of motion for a multi-body connected system, using the formula $a = \frac{m_3 g - m_1 g}{m_1 + m_2 + m_3}$.

From the image given below, find the acceleration $a$ of the system consisting of a $10 \, \text{kg}$ hanging block, a $20 \, \text{kg}$ block on a smooth table, and a $20 \, \text{kg}$ hanging block connected over two fixed pulleys.

A

$1 \, \text{m/s}^2$

B

$2 \, \text{m/s}^2$

C

$3 \, \text{m/s}^2$

D

$4 \, \text{m/s}^2$

Question 33: Palette 33/105

DPT MCQ

Newton second law of motion for an equivalent Atwood machine setup, using the formulas $a = \frac{m_1 g - m_2 g}{m_1 + m_2}$ and $m_1 g - T = m_1 a$.

From the image given below, find the acceleration $a$ of the system and the string tension $T$ when a $3 \, \text{kg}$ hanging block and a $2 \, \text{kg}$ hanging block are connected over pulleys in an arrangement.

A

$a = 1 \, \text{m/s}^2, T = 20 \, \text{N}$

B

$a = 2 \, \text{m/s}^2, T = 24 \, \text{N}$

C

$a = 3 \, \text{m/s}^2, T = 18 \, \text{N}$

D

$a = 4 \, \text{m/s}^2, T = 22 \, \text{N}$

Question 34: Palette 34/105

DPT MCQ

Newton second law for a system with an external opposing force on the horizontal mass, using the acceleration formula $a = \frac{m_1 g - F_0}{m_1 + m_2}$.

From the image given below, find the acceleration $a$ of the system where a $4 \, \text{kg}$ block on a smooth table has an opposing force $F_0 = 20 \, \text{N}$ and is connected to a hanging $6 \, \text{kg}$ block.

A

$a = 1 \, \text{m/s}^2$

B

$a = 2 \, \text{m/s}^2$

C

$a = 3 \, \text{m/s}^2$

D

$a = 4 \, \text{m/s}^2$

Question 35: Palette 35/105

DPT MCQ

Newton second law of motion for a complex connected multi-body system involving an inclined plane and multiple hanging masses, using the formula $a = \frac{m_3 g - m_2 g \sin\theta - m_1 g}{m_1 + m_2 + m_3}$.

From the image given below, find the acceleration $a$ of the system consisting of a $20 \, \text{kg}$ hanging block, a $10 \, \text{kg}$ block on a $30^\circ$ inclined plane, and a $4 \, \text{kg}$ hanging block connected over pulleys.

A

$a = 2.15 \, \text{m/s}^2$

B

$a = 3.24 \, \text{m/s}^2$

C

$a = 4.50 \, \text{m/s}^2$

D

$a = 5.10 \, \text{m/s}^2$

Question 36: Palette 36/105

DPT MCQ

Newton second law of motion for masses on two inclined planes, using the formula $a = \frac{m_2 g \sin\theta_2 - m_1 g \sin\theta_1}{m_1 + m_2}$.

From the image given below, find the acceleration $a$ of the system where two $10 \, \text{kg}$ blocks are placed on opposite inclined planes of $37^\circ$ and $53^\circ$ connected by a string over a pulley at the top.

A

$a = 1 \, \text{m/s}^2$

B

$a = 2 \, \text{m/s}^2$

C

$a = 3 \, \text{m/s}^2$

D

$a = 4 \, \text{m/s}^2$

Question 37: Palette 37/105

DPT MCQ

Newton second law of motion for a multi-mass system on a horizontal surface connected to a hanging mass, using the formulas $a = \frac{m_{\text{hanging}} g}{\sum m_i}$ and $T = m_{\text{effective}} a$.

From the image given below, find the acceleration $a$ of the system and the string tensions $T_1$ and $T_2$ for blocks of masses $5 \, \text{kg}$, $8 \, \text{kg}$, and $2 \, \text{kg}$ on a smooth table connected to a hanging $10 \, \text{kg}$ block.

A

$a = 2 \, \text{m/s}^2, T_1 = 10 \, \text{N}, T_2 = 26 \, \text{N}$

B

$a = 4 \, \text{m/s}^2, T_1 = 20 \, \text{N}, T_2 = 52 \, \text{N}$

C

$a = 5 \, \text{m/s}^2, T_1 = 25 \, \text{N}, T_2 = 65 \, \text{N}$

D

$a = 6 \, \text{m/s}^2, T_1 = 30 \, \text{N}, T_2 = 78 \, \text{N}$

Question 38: Palette 38/105

DPT MCQ

Newton second law for system acceleration $a = \frac{m_{\text{hanging}} g}{m_1 + m_2}$ followed by kinematics equation of motion $s = ut + \frac{1}{2} a t^2$.

From the image given below, find the time when the $6 \, \text{kg}$ block strikes the ground if it starts from rest at a height of $10 \, \text{m}$ connected to a $4 \, \text{kg}$ block on a smooth table.

A

$t = \sqrt{\frac{5}{3}} \, \text{s}$

B

$t = \sqrt{\frac{10}{3}} \, \text{s}$

C

$t = \sqrt{\frac{20}{3}} \, \text{s}$

D

$t = \sqrt{\frac{40}{3}} \, \text{s}$

Question 39: Palette 39/105

DPT MCQ

The work done by tension forces in a string-pulley system is zero, or the sum of power developed by tension is zero ($\sum \vec{T} \cdot \vec{a} = 0$).

Find the acceleration $a_3$ of the bottom block in the given pulley system where the acceleration of the first block is $a_1 = 4\text{ m/s}^2$ to the left and the second block is $a_2 = 6\text{ m/s}^2$ to the left.

A

$3\text{ m/s}^2$

B

$5\text{ m/s}^2$

C

$4\text{ m/s}^2$

D

$6\text{ m/s}^2$

Question 40: Palette 40/105

DPT MCQ

The net work done by internal tension forces in a string-pulley system over any displacement or power developed is zero, expressed by the formula $\sum T \cdot a = 0$.

Find $a'$ in the following string-pulley system where a block moves upwards with acceleration $a = 2\text{ m/s}^2$ on the left and another block moves downwards with acceleration $a = 8\text{ m/s}^2$ on the right.

A

$4\text{ m/s}^2$

B

$3\text{ m/s}^2$

C

$5\text{ m/s}^2$

D

$2\text{ m/s}^2$

Question 41: Palette 41/105

DPT MCQ

The work done by virtual tension forces in a system of connected bodies is zero, represented by the work-energy constraint formula $\sum \vec{T} \cdot \vec{a} = 0$.

Find the acceleration $a'$ of the block on the inclined plane in the given pulley system where the small block moves downwards with acceleration $a = 2\text{ m/s}^2$.

A

$1\text{ m/s}^2$

B

$2\text{ m/s}^2$

C

$3\text{ m/s}^2$

D

$4\text{ m/s}^2$

Question 42: Palette 42/105

DPT MCQ

The work done by tension forces in an ideal string-pulley system is zero, given by the virtual work constraint formula $\sum \vec{T} \cdot \vec{a} = 0$.

Find $a_2$ in the given pulley system where block 1 moves upwards with acceleration $a_1 = 4\text{ m/s}^2$ and block 2 moves downwards with acceleration $a_2$.

A

$1\text{ m/s}^2$

B

$2\text{ m/s}^2$

C

$3\text{ m/s}^2$

D

$4\text{ m/s}^2$

Question 43: Palette 43/105

DPT MCQ

The work done by tension forces in an ideal string-pulley system is zero, given by the virtual work constraint formula $\sum \vec{T} \cdot \vec{a} = 0$.

Find $a'$ in the given string-pulley system where the first block moves upwards with acceleration $a_1 = 2\text{ m/s}^2$, the second block moves upwards with acceleration $a_2 = 3\text{ m/s}^2$, and the third block moves downwards with acceleration $a_3 = 5\text{ m/s}^2$.

A

$1\text{ m/s}^2$

B

$4\text{ m/s}^2$

C

$2\text{ m/s}^2$

D

$3\text{ m/s}^2$

Question 44: Palette 44/105

DPT MCQ

The work done by tension forces in an ideal string-pulley system is zero, given by the virtual work constraint formula $\sum \vec{T} \cdot \vec{a} = 0$.

Find $a'$ in the given figure where the left block moves upwards with acceleration $a_1 = 6\text{ m/s}^2$ and the green platform moves downwards with acceleration $a'$.

A

$4\text{ m/s}^2$

B

$3\text{ m/s}^2$

C

$2\text{ m/s}^2$

D

$1\text{ m/s}^2$

Question 45: Palette 45/105

DPT MCQ

The work done by virtual tension forces in a system of connected bodies is zero, represented by the work-energy constraint formula $\sum \vec{T} \cdot \vec{a} = 0$.

Find the acceleration $a_3$ of the platform in the given string-pulley system where block 1 moves up the incline with acceleration $a_1 = 4\text{ m/s}^2$ and block 2 moves downwards with acceleration $a_2 = 6\text{ m/s}^2$.

A

$-\frac{10}{3}\text{ m/s}^2$

B

$-\frac{14}{3}\text{ m/s}^2$

C

$-\frac{12}{3}\text{ m/s}^2$

D

$-\frac{8}{3}\text{ m/s}^2$

Question 46: Palette 46/105

DPT MCQ

The work done by string tension in a connected system is zero, defined by the power dot product formula $\vec{T}_1 \cdot \vec{a}_1 + \vec{T}_2 \cdot \vec{a}_2 = 0$.

Find $a_2$ in the given system where block 1 moves horizontally to the right with acceleration $a_1 = 10\text{ m/s}^2$ and tension $T$ makes an angle of $37^\circ$ with the horizontal.

A

$6\text{ m/s}^2$

B

$8\text{ m/s}^2$

C

$10\text{ m/s}^2$

D

$5\text{ m/s}^2$

Question 47: Palette 47/105

DPT MCQ

The work done by string tension in a connected system is zero, defined by the power dot product formula $\vec{T}_1 \cdot \vec{a}_1 + \vec{T}_2 \cdot \vec{a}_2 = 0$.

Find $a_2$ in the given system where a ring slides along a horizontal rod with acceleration $a_1 = 10\text{ m/s}^2$ and the string attached to it makes an angle of $37^\circ$ with the rod.

A

$6\text{ m/s}^2$

B

$10\text{ m/s}^2$

C

$8\text{ m/s}^2$

D

$4\text{ m/s}^2$

Question 48: Palette 48/105

DPT MCQ

The net work done by internal tension forces (or power developed by tension) in an ideal string-pulley system is zero, defined by the formula $\sum \vec{T} \cdot \vec{v} = 0$.

In the figure, find the velocity of block B, if velocity of A is $V_0$ in downward direction?

A

$\frac{V_0}{\cos \theta}$

B

$\frac{V_0}{1 + \cos \theta}$

C

$V_0 (1 + \cos \theta)$

D

$\frac{V_0}{\sin \theta}$

Question 49: Palette 49/105

DPT MCQ

The net work done by internal tension forces in an ideal string-pulley system is zero, defined by the formula $\sum \vec{T} \cdot \vec{a} = 0$.

Find $a_3$ in the given string-pulley system where the first block moves upwards with acceleration $a_1 = 2\text{ m/s}^2$, the second block moves upwards with acceleration $a_2 = 4\text{ m/s}^2$, and the third block has acceleration $a_3$.

A

$-2\text{ m/s}^2$

B

$-6\text{ m/s}^2$

C

$-4\text{ m/s}^2$

D

$-3\text{ m/s}^2$

Question 50: Palette 50/105

DPT MCQ

The work done by tension forces in an ideal string-pulley system is zero, defined by the virtual work constraint formula $\sum \vec{T} \cdot \vec{a} = 0$.

Find the acceleration $a'$ of the block on the horizontal surface in the given pulley system where the bottom platform moves downwards with acceleration $a = 6\text{ m/s}^2$.

A

$12\text{ m/s}^2$

B

$16\text{ m/s}^2$

C

$18\text{ m/s}^2$

D

$24\text{ m/s}^2$

Question 51: Palette 51/105

DPT MCQ

The work done by internal tension forces in an ideal string-pulley system is zero, defined by the virtual work constraint formula $\sum \vec{T} \cdot \vec{a} = 0$.

Find $a'$ in the given string-pulley system where the first block moves upwards with acceleration $a' = ?$, the second block moves upwards with acceleration $a = 2\text{ m/s}^2$, the third block moves downwards with acceleration $a = 6\text{ m/s}^2$, and the fourth block/platform moves downwards with acceleration $a = 4\text{ m/s}^2$.

A

$15\text{ m/s}^2$

B

$20\text{ m/s}^2$

C

$18\text{ m/s}^2$

D

$24\text{ m/s}^2$

Question 52: Palette 52/105

DPT MCQ

Constraint relation from virtual work formula $\sum \vec{T} \cdot \vec{a} = 0$ combined with Newton's second law $\sum F = m a$.

Find the acceleration of each block in the given pulley system containing a $4\text{ kg}$ block and a $10\text{ kg}$ block (taking $g = 10\text{ m/s}^2$).

A

$a_1 = \frac{5}{13}\text{ m/s}^2, a_2 = \frac{10}{13}\text{ m/s}^2$

B

$a_1 = \frac{10}{13}\text{ m/s}^2, a_2 = \frac{20}{13}\text{ m/s}^2$

C

$a_1 = \frac{15}{13}\text{ m/s}^2, a_2 = \frac{30}{13}\text{ m/s}^2$

D

$a_1 = \frac{20}{13}\text{ m/s}^2, a_2 = \frac{40}{13}\text{ m/s}^2$

Question 53: Palette 53/105

DPT MCQ

The work done by virtual tension forces gives the constraint relation $\sum \vec{T} \cdot \vec{a} = 0$, combined with Newton's second law $a = \frac{F_{\text{net}}}{m}$.

Find the acceleration of each block in the given pulley system containing a $4\text{ kg}$ block and a $6\text{ kg}$ block (taking $g = 10\text{ m/s}^2$).

A

$a_1 = \frac{20}{7}\text{ m/s}^2, a_2 = -\frac{40}{7}\text{ m/s}^2$

B

$a_1 = \frac{10}{7}\text{ m/s}^2, a_2 = -\frac{20}{7}\text{ m/s}^2$

C

$a_1 = \frac{30}{7}\text{ m/s}^2, a_2 = -\frac{60}{7}\text{ m/s}^2$

D

$a_1 = \frac{15}{7}\text{ m/s}^2, a_2 = -\frac{30}{7}\text{ m/s}^2$

Question 54: Palette 54/105

DPT MCQ

The work done by virtual tension forces gives the constraint relation $\sum \vec{T} \cdot \vec{a} = 0$, combined with Newton's second law $a = \frac{F_{\text{net}}}{m}$.

Find the acceleration of each block in the given pulley system containing a $4\text{ kg}$ block and an $8\text{ kg}$ block (taking $g = 10\text{ m/s}^2$).

A

$a_1 = \frac{10}{11}\text{ m/s}^2, a_2 = -\frac{30}{11}\text{ m/s}^2$

B

$a_1 = \frac{15}{11}\text{ m/s}^2, a_2 = -\frac{45}{11}\text{ m/s}^2$

C

$a_1 = \frac{20}{11}\text{ m/s}^2, a_2 = -\frac{60}{11}\text{ m/s}^2$

D

$a_1 = \frac{5}{11}\text{ m/s}^2, a_2 = -\frac{15}{11}\text{ m/s}^2$

Question 55: Palette 55/105

DPT MCQ

The work done by virtual tension forces gives the string constraint equation $\sum \vec{T} \cdot \vec{a} = 0$, combined with Newton's second law $a = \frac{F_{\text{net}}}{m}$.

Find the acceleration of each block in the given pulley system containing blocks of masses $2\text{ kg}$, $4\text{ kg}$, $8\text{ kg}$, and $16\text{ kg}$ (taking $g = 10\text{ m/s}^2$).

A

$a_1 = \frac{290}{99}\text{ m/s}^2, a_2 = \frac{290}{99}\text{ m/s}^2, a_3 = -\frac{830}{99}\text{ m/s}^2, a_4 = -\frac{910}{99}\text{ m/s}^2$

B

$a_1 = \frac{145}{99}\text{ m/s}^2, a_2 = \frac{145}{99}\text{ m/s}^2, a_3 = -\frac{415}{99}\text{ m/s}^2, a_4 = -\frac{455}{99}\text{ m/s}^2$

C

$a_1 = \frac{580}{99}\text{ m/s}^2, a_2 = \frac{580}{99}\text{ m/s}^2, a_3 = -\frac{1660}{99}\text{ m/s}^2, a_4 = -\frac{1820}{99}\text{ m/s}^2$

D

$a_1 = \frac{100}{99}\text{ m/s}^2, a_2 = \frac{100}{99}\text{ m/s}^2, a_3 = -\frac{300}{99}\text{ m/s}^2, a_4 = -\frac{400}{99}\text{ m/s}^2$

Question 56: Palette 56/105

DPT MCQ

The work done by virtual tension forces gives the string constraint relation $\sum \vec{T} \cdot \vec{a} = 0$, combined with Newton's second law $a = \frac{F_{\text{net}}}{m}$.

Find the acceleration of each block in the given pulley system containing blocks of masses $1\text{ kg}$, $2\text{ kg}$, and $3\text{ kg}$ (taking $g = 10\text{ m/s}^2$).

A

$a_1 = \frac{95}{29}\text{ m/s}^2, a_2 = -\frac{85}{29}\text{ m/s}^2, a_3 = -\frac{105}{29}\text{ m/s}^2$

B

$a_1 = \frac{190}{29}\text{ m/s}^2, a_2 = -\frac{170}{29}\text{ m/s}^2, a_3 = -\frac{210}{29}\text{ m/s}^2$

C

$a_1 = \frac{380}{29}\text{ m/s}^2, a_2 = -\frac{340}{29}\text{ m/s}^2, a_3 = -\frac{420}{29}\text{ m/s}^2$

D

$a_1 = \frac{120}{29}\text{ m/s}^2, a_2 = -\frac{140}{29}\text{ m/s}^2, a_3 = -\frac{160}{29}\text{ m/s}^2$

Question 57: Palette 57/105

DPT MCQ

The work done by tension forces in an ideal string-pulley system provides the constraint relation $\sum \vec{T} \cdot \vec{a} = 0$, combined with Newton's second law $\sum F = m a$.

Find the acceleration of mass $m$ in the given string-pulley system where block $2m$ rests on a smooth horizontal surface and block $m$ hangs vertically.

A

$-\frac{g}{3}$

B

$-\frac{2g}{3}$

C

$-\frac{g}{2}$

D

$-\frac{4g}{3}$

Question 58: Palette 58/105

DPT MCQ

The work done by virtual tension forces gives the string constraint relation $\sum \vec{T} \cdot \vec{a} = 0$, combined with Newton's second law $a = \frac{F_{\text{net}}}{m}$.

Find the acceleration of both blocks in the given string-pulley system where block of mass $m$ is on an inclined plane of angle $30^\circ$ and block of mass $2m$ is hanging vertically.

A

$a_1 = \frac{g}{3}, a_2 = -\frac{g}{6}$

B

$a_1 = \frac{g}{6}, a_2 = -\frac{g}{12}$

C

$a_1 = \frac{g}{2}, a_2 = -\frac{g}{4}$

D

$a_1 = \frac{2g}{3}, a_2 = -\frac{g}{3}$

Question 59: Palette 59/105

DPT MCQ

Using string constraint $\sum \vec{T} \cdot \vec{a} = 0$ along with Newton's second law $\sum F = ma$.

In the given figure, find the acceleration of block $B$ if all surfaces are frictionless.

A

$\frac{F}{7m}$

B

$\frac{F}{14m}$

C

$\frac{2F}{7m}$

D

$\frac{F}{21m}$

Question 60: Palette 60/105

DPT MCQ

The work done per unit time by virtual tension forces in a string system is zero, given by $\sum \vec{T} \cdot \vec{V} = 0$.

Find out the velocity of block $E$ as shown in the figure.

A

$\frac{31}{2}\text{ ms}^{-1}$

B

$\frac{29}{2}\text{ ms}^{-1}$

C

$\frac{33}{2}\text{ ms}^{-1}$

D

$\frac{27}{2}\text{ ms}^{-1}$

Question 61: Palette 61/105

DPT MCQ

Concept

+

To find the acceleration and friction, we first calculate the maximum static friction $(f_{s})_{max} = \mu_s N$ and kinetic friction $f_k = \mu_k N$, where $N = mg$. If the applied force $F \le (f_{s})_{max}$, the block remains stationary, acceleration $a = 0$, and the static friction $f = F$. If $F > (f_{s})_{max}$, the block moves, and the kinetic friction $f = f_k$ applies.

In the given figure, find the value of acceleration and friction if $F = 10 N$, given mass of the block is $10 kg$, $\mu_s = 0.4$, and $\mu_k = 0.3$.

A

$a = 1 m/s^2$, $f = 10 N$

B

$a = 0$, $f = 10 N$

C

$a = 0.5 m/s^2$, $f = 30 N$

D

$a = 0$, $f = 40 N$

Question 62: Palette 62/105

DPT MCQ

Concept

+

To find the acceleration and friction, we first calculate the maximum static friction $(f_{s})_{max} = \mu_s N$ and kinetic friction $f_k = \mu_k N$, where $N = mg$. If the applied force $F \le (f_{s})_{max}$, the block remains stationary, acceleration $a = 0$, and static friction $f = F$. If $F > (f_{s})_{max}$, the block moves, and $a = \frac{F - f_k}{m}$ with $f = f_k$.

In the given figure, find the value of acceleration and friction if $F = 20 N$, given mass of the block is $10 kg$, $\mu_s = 0.4$, and $\mu_k = 0.3$.

A

$a = 0$, $f = 20 N$

B

$a = 1 m/s^2$, $f = 30 N$

C

$a = 0.5 m/s^2$, $f = 20 N$

D

$a = 2 m/s^2$, $f = 40 N$

Question 63: Palette 63/105

DPT MCQ

Concept

+

To find the acceleration and friction, we first calculate the maximum static friction $(f_{s})_{max} = \mu_s N$ and kinetic friction $f_k = \mu_k N$, where $N = mg$. If the applied force $F \le (f_{s})_{max}$, the block remains stationary, acceleration $a = 0$, and static friction $f = F$. If $F > (f_{s})_{max}$, the block moves, and $a = \frac{F - f_k}{m}$ with $f = f_k$.

In the given figure, find the value of acceleration and friction if $F = 35 N$, given mass of the block is $10 kg$, $\mu_s = 0.4$, and $\mu_k = 0.3$.

A

$a = 0.5 m/s^2$, $f = 30 N$

B

$a = 0$, $f = 35 N$

C

$a = 1 m/s^2$, $f = 35 N$

D

$a = 0$, $f = 40 N$

Question 64: Palette 64/105

DPT MCQ

Concept

+

To find the acceleration and friction, we first calculate the maximum static friction $(f_{s})_{max} = \mu_s N$ and kinetic friction $f_k = \mu_k N$, where $N = mg$. If the applied force $F \le (f_{s})_{max}$, the block remains stationary, acceleration $a = 0$, and static friction $f = F$. If $F > (f_{s})_{max}$, the block moves, and $a = \frac{F - f_k}{m}$ with $f = f_k$.

In the given figure, find the value of acceleration and friction if $F = 39.99 N$, given mass of the block is $10 kg$, $\mu_s = 0.4$, and $\mu_k = 0.3$.

A

$a = 0.99 m/s^2$, $f = 30 N$

B

$a = 0$, $f = 30 N$

C

$a = 0$, $f = 39.99 N$

D

$a = 1 m/s^2$, $f = 39.99 N$

Question 65: Palette 65/105

DPT MCQ

Concept

+

To find the acceleration when the applied force exceeds the maximum static friction $(f_{s})_{max}$, the block starts moving and we use Newton's second law: $a = \frac{F - f_k}{m}$, where $f_k = \mu_k N$ is the kinetic friction.

In the given figure, find the value of acceleration if $F = 40.01 N$, given mass of the block is $10 kg$, $\mu_s = 0.4$, and $\mu_k = 0.3$.

A

$a = 1.001 m/s^2$

B

$a = 2 m/s^2$

C

$a = 0.5 m/s^2$

D

$a = 0$

Question 66: Palette 66/105

DPT MCQ

Concept

+

To find the acceleration when the applied force exceeds the maximum static friction $(f_{s})_{max}$, the block moves with kinetic friction and the acceleration is given by $a = \frac{F - f_k}{m}$.

In the given figure, find the value of acceleration if $F = 50 N$, given mass of the block is $10 kg$, $\mu_s = 0.4$, and $\mu_k = 0.3$.

A

$a = 1 m/s^2$

B

$a = 2 m/s^2$

C

$a = 1.5 m/s^2$

D

$a = 0$

Question 67: Palette 67/105

DPT MCQ

Concept

+

The normal force on the block is $N = mg + 100 N = 200 N$. The maximum static friction is $(f_s)_{max} = \mu N = 0.1 \times 200 N = 20 N$. The net horizontal force is given by the difference between $F_1$ and $F_2$. If the net applied force is less than or equal to $(f_s)_{max}$, the block remains at rest and the static friction equals the net applied force in magnitude.

Find the frictional force acting on the block for Case-I when $F_1 = 0$ and $F_2 = 5 N$, given the mass of the block is $10 kg$ and $\mu = 0.1$ with $g = 10 m/s^2$.

A

$f = 5 N$ towards left

B

$f = 10 N$ towards right

C

$f = 20 N$ towards left

D

$f = 0$

Question 68: Palette 68/105

DPT MCQ

Concept

+

The normal force is $N = 200 N$ and $(f_s)_{max} = 20 N$. When the magnitude of the net applied force $\vert{}F_2 - F_1\vert{}$ is less than $(f_s)_{max}$, the static friction matches the net applied force to keep the block stationary.

Find the frictional force acting on the block for Case-I when $F_1 = 0$ and $F_2 = 15 N$, given the mass of the block is $10 kg$ and $\mu = 0.1$ with $g = 10 m/s^2$.

A

$f = 10 N$ towards left

B

$f = 15 N$ towards left

C

$f = 20 N$ towards right

D

$f = 0$

Question 69: Palette 69/105

DPT MCQ

Concept

+

When the net applied force $\vert{}F_2 - F_1\vert{}$ exceeds the maximum static friction $(f_s)_{max} = \mu N$, the block moves and kinetic friction $f_k = \mu N$ comes into play.

Find the frictional force acting on the block for Case-I when $F_1 = 0$ and $F_2 = 25 N$, given the mass of the block is $10 kg$ and $\mu = 0.1$ with $g = 10 m/s^2$.

A

$f = 25 N$ towards left

B

$f = 20 N$ towards left

C

$f = 15 N$ towards right

D

$f = 0$

Question 70: Palette 70/105

DPT MCQ

Concept

+

The net applied force is the absolute difference $\vert{}F_2 - F_1\vert{}$. If this is less than or equal to the maximum static friction $(f_s)_{max}$, the static friction equals the net force.

Find the frictional force acting on the block for Case-II when $F_1 = 20 N$ and $F_2 = 10 N$, given the mass of the block is $10 kg$ and $\mu = 0.1$ with $g = 10 m/s^2$.

A

$f = 20 N$ towards right

B

$f = 10 N$ towards right

C

$f = 5 N$ towards left

D

$f = 0$

Question 71: Palette 71/105

DPT MCQ

Concept

+

When opposing forces are equal in magnitude, the net applied force is zero, resulting in zero friction and no motion.

Find the frictional force acting on the block for Case-II when $F_1 = 20 N$ and $F_2 = 20 N$, given the mass of the block is $10 kg$ and $\mu = 0.1$ with $g = 10 m/s^2$.

A

$f = 20 N$

B

$f = 10 N$

C

$f = 0$

D

$f = 40 N$

Question 72: Palette 72/105

DPT MCQ

Concept

+

When the net applied force $\vert{}F_2 - F_1\vert{}$ exceeds $(f_s)_{max} = 20 N$, the block moves and kinetic friction acts opposite to the direction of motion.

Find the frictional force acting on the block for Case-II when $F_1 = 20 N$ and $F_2 = 42 N$, given the mass of the block is $10 kg$ and $\mu = 0.1$ with $g = 10 m/s^2$.

A

$f = 42 N$ towards left

B

$f = 20 N$ towards left

C

$f = 10 N$ towards right

D

$f = 0$

Question 73: Palette 73/105

DPT MCQ

Concept

+

When a force $F$ is applied at an angle $\theta$ ($37^\circ$) to the horizontal, we resolve it into components: horizontal component $F \cos 37^\circ$ and vertical component $F \sin 37^\circ$. The normal force is balanced as $N + F \sin 37^\circ = mg$, so $N = mg - F \sin 37^\circ$. The maximum static friction is $(f_{s})_{max} = \mu_s N$. If the horizontal driving force $F \cos 37^\circ \le (f_{s})_{max}$, the block remains stationary with $a = 0$ and static friction equal to $F \cos 37^\circ$.

Find the frictional force and acceleration of the block for case (i) when $F = 10 N$, given the mass of the block is $10 kg$, $\mu_s = 0.3$, and $\mu_k = 0.2$ with an angle of $37^\circ$.

A

$a = 0$, $f = 8 N$

B

$a = 1 m/s^2$, $f = 28.2 N$

C

$a = 0.5 m/s^2$, $f = 10 N$

D

$a = 0$, $f = 15 N$

Question 74: Palette 74/105

DPT MCQ

Concept

+

For an inclined pulling force, $N = mg - F \sin 37^\circ$ and $(f_{s})_{max} = \mu_s N$. If the horizontal driving force $F \cos 37^\circ > (f_{s})_{max}$, the block moves, and the kinetic friction $f_k = \mu_k N$ acts on it, with acceleration calculated using Newton's second law: $a = \frac{F \cos 37^\circ - f_k}{m}$.

Find the frictional force and acceleration of the block for case (ii) when $F = 50 N$, given the mass of the block is $10 kg$, $\mu_s = 0.3$, and $\mu_k = 0.2$ with an angle of $37^\circ$.

A

$a = 0$, $f = 50 N$

B

$a = 2.6 m/s^2$, $f = 14 N$

C

$a = 2.5 m/s^2$, $f = 20 N$

D

$a = 0$, $f = 28.2 N$

Question 75: Palette 75/105

DPT MCQ

Concept

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When a force $F$ pushes a block downwards at an angle $\theta$ ($37^\circ$), the normal force is $N = mg + F \sin 37^\circ$. The maximum static friction is $(f_{s})_{max} = \mu N$. If the horizontal driving force $F \cos 37^\circ > (f_{s})_{max}$, the block accelerates, and the acceleration is given by $a = \frac{F \cos 37^\circ - (f_{s})_{max}}{m}$.

Find the acceleration of the block if $F = 100 N$, given the mass of the block is $10 kg$, $\mu = 0.2$, and the force acts at an angle of $37^\circ$ downwards.

A

$a = 4.8 m/s^2$

B

$a = 2.5 m/s^2$

C

$a = 6.0 m/s^2$

D

$a = 3.2 m/s^2$

Question 76: Palette 76/105

DPT MCQ

Concept

+

When a force $F$ pushes a block downwards at an angle $\theta$ ($37^\circ$), the normal force is $N = mg + F \sin 37^\circ$. The maximum static friction is $(f_{s})_{max} = \mu N$. If the horizontal driving force $F \cos 37^\circ > (f_{s})_{max}$, the block accelerates, and the acceleration is given by $a = \frac{F \cos 37^\circ - (f_{s})_{max}}{m}$.

Find the acceleration of the block if $F = 200 N$, given the mass of the block is $10 kg$, $\mu = 0.2$, and the force acts at an angle of $37^\circ$ downwards.

A

$a = 11.2 m/s^2$

B

$a = 12.8 m/s^2$

C

$a = 10.5 m/s^2$

D

$a = 9.6 m/s^2$

Question 77: Palette 77/105

DPT MCQ

Concept

+

When a block of mass $m_1$ is connected by a string over a pulley to a hanging mass $m_2$, the normal force on $m_1$ is $N = m_1 g$. The maximum static friction is $(f_{s})_{max} = \mu_s N$ and the kinetic friction is $f_k = \mu_k N$. If the driving force $m_2 g > (f_{s})_{max}$, the system accelerates such that $a = \frac{m_2 g - f_k}{m_1 + m_2}$ and the tension is $T = m_2 g - m_2 a$.

Find the acceleration of the $10 kg$ block and the tension in the string for the system shown in the figure where a $5 kg$ mass hangs vertically, given $\mu_s = 0.4$ and $\mu_k = 0.3$.

A

$a = \frac{4}{3} m/s^2$, $T = \frac{130}{3} N$

B

$a = 2 m/s^2$, $T = 40 N$

C

$a = \frac{5}{3} m/s^2$, $T = \frac{110}{3} N$

D

$a = 1 m/s^2$, $T = 45 N$

Question 78: Palette 78/105

DPT MCQ

Concept

+

For a block pressed against a vertical wall, the downward gravitational force $mg$ is balanced by the upward static friction force $f$. The normal reaction force $N$ provided by the wall is equal to the applied perpendicular force $F$, so $N = F$. The maximum static friction force is $f_{max} = \mu N = \mu F$. To prevent the block from slipping, $f_{max} \ge mg$, which gives the minimum force $F_{min} = \frac{mg}{\mu}$.

A block of 1 kg is stopped against a wall by applying a force F perpendicular to the wall. If $\mu = 0.2$ then minimum value of F will be

A

980 N

B

49 N

C

98 N

D

490 N

Question 79: Palette 79/105

DPT MCQ

Concept

+

The coefficient of static friction ($\mu_s$) is the ratio of the maximum static frictional force ($F_s$, the minimum horizontal force needed to set the block in motion) to the normal force ($N$) exerted by the surface.
Formula: $\mu_s = \frac{F_s}{N} = \frac{F_s}{mg}$

A 20 kg block is initially at rest on a rough horizontal surface. A horizontal force of 75 N is required to set the block in motion. After it is in motion, a horizontal force of 60 N is required to keep the block moving with constant speed. The coefficient of static friction is

A

0.38

B

0.44

C

0.52

D

0.60

Question 80: Palette 80/105

DPT MCQ

Concept

+

Static friction is a self-adjusting force. As long as the applied horizontal force $F$ is less than or equal to the maximum limiting static friction $f_{max} = \mu_s N = \mu_s mg$, the frictional force $f$ acting on the body is equal in magnitude to the applied force $F$.
Formula: $f = F$ when $F \le f_{max}$, where $f_{max} = \mu_s mg$.

A block of mass 2 kg is kept on the floor. The coefficient of static friction is 0.4. If a force F of 2.5 Newtons is applied on the block as shown in the figure, the frictional force between the block and the floor will be

A

2.5 N

B

5 N

C

7.84 N

D

10 N

Question 81: Palette 81/105

DPT MCQ

Concept

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The maximum static friction available at the contact surface is given by the product of the coefficient of static friction and the normal reaction force from the floor. Since the floor supports the entire vertical weight of the ladder, the normal reaction force equals the weight of the ladder.
Formula: $f_{max} = \mu N = \mu W$

If a ladder weighing 250 N is placed against a smooth vertical wall having coefficient of friction between it and floor is 0.3, then what is the maximum force of friction available at the point of contact between the ladder and the floor

A

25 N

B

50 N

C

35 N

D

75 N

Question 82: Palette 82/105

DPT MCQ

Concept

+

For a body pressed against a vertical wall, the normal reaction force $N$ equals the applied horizontal pressing force $F$. The maximum possible static frictional force is $f_{max} = \mu N$. To hold the body in equilibrium, the upward frictional force $f$ balances the downward gravitational weight $mg$, provided that $mg \le f_{max}$. If $mg \le f_{max}$, the actual frictional force acting on the body is equal to its weight $mg$.
Formula: $N = F$, $f_{max} = \mu N$, and $f = mg$ (when $mg \le f_{max}$).

A body of mass 2 kg is kept by pressing to a vertical wall by a force of 100 N. The coefficient of friction between wall and body is 0.3. Then the frictional force is equal to

A

6 N

B

20 N

C

600 N

D

700 N

Question 83: Palette 83/105

DPT MCQ

Concept

+

When a body is placed on a rough horizontal surface and a horizontal force $F$ is applied, the net force causing acceleration is the applied force minus the kinetic frictional force $f_k$.
Formulas involved:
Normal reaction force: $N = mg$
Kinetic frictional force: $f_k = \mu_k N = \mu_k mg$
Net force: $F_{net} = F - f_k = ma$
Acceleration: $a = \frac{F - \mu_k mg}{m}$

A block of mass 5 kg is on a rough horizontal surface and is at rest. Now a force of 24 N is imparted to it with negligible impulse. If the coefficient of kinetic friction is 0.4 and $g = 9.8\text{ m/s}^2$, then the acceleration of the block is

A

$0.26\text{ m/s}^2$

B

$0.39\text{ m/s}^2$

C

$0.69\text{ m/s}^2$

D

$0.88\text{ m/s}^2$

Question 84: Palette 84/105

DPT MCQ

Concept

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When a body moves with uniform velocity on a rough horizontal surface, the external force applied is equal to the kinetic friction force $f_k = \mu mg$. The work done against friction is converted into heat energy.
Formulas involved:
Kinetic friction force: $f = \mu mg$
Distance covered in time $t$: $s = v \times t$
Work done against friction: $W = f \times s = \mu mg \times v \times t$
Heat generated in calories: $H = \frac{W}{J}$

A body of mass 2 kg is being dragged with uniform velocity of 2 m/s on a rough horizontal plane. The coefficient of friction between the body and the surface is 0.20. The amount of heat generated in 5 sec is ($J = 4.2\text{ joule/cal}$ and $g = 9.8\text{ m/s}^2$)

A

9.33 cal

B

10.21 cal

C

12.67 cal

D

13.34 cal

Question 85: Palette 85/105

DPT MCQ

Concept

+

When an external force $F$ is applied at an angle $\theta$ to the horizontal on a block on a rough horizontal surface, it resolves into horizontal component $F \cos\theta$ and vertical component $F \sin\theta$. The normal reaction force $N$ is reduced by the vertical component: $N = Mg - F \sin\theta$. The kinetic friction opposing motion is $f_k = \mu N$. The net force along the horizontal direction causes acceleration $a$.
Formulas involved:
Normal reaction: $N = Mg - F \sin\theta$
Friction force: $f_k = \mu N = \mu (Mg - F \sin\theta)$
Acceleration: $a = \frac{F \cos\theta - f_k}{M} = \frac{F \cos\theta - \mu(Mg - F \sin\theta)}{M}$

A block of mass $M = 5\text{ kg}$ is resting on a rough horizontal surface for which the coefficient of friction is 0.2. When a force $F = 40\text{ N}$ is applied, the acceleration of the block will be ($g = 10\text{ m/s}^2$)

A

$5.73\text{ m/sec}^2$

B

$8.0\text{ m/sec}^2$

C

$3.17\text{ m/sec}^2$

D

$10.0\text{ m/sec}^2$

Question 86: Palette 86/105

DPT MCQ

Concept

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When a body slides down a rough inclined plane, the downward force along the incline due to gravity is $mg \sin\theta$. The normal reaction force perpendicular to the inclined surface is $N = mg \cos\theta$, which creates an opposing kinetic frictional force $f_k = \mu N = \mu mg \cos\theta$. The net force accelerating the body down the incline is $F_{net} = mg \sin\theta - \mu mg \cos\theta$.
Formulas involved:
Net force: $F_{net} = m g (\sin\theta - \mu \cos\theta)$
Acceleration: $a = \frac{F_{net}}{m} = g (\sin\theta - \mu \cos\theta)$

When a body is placed on a rough plane inclined at an angle $\theta$ to the horizontal, its acceleration is

A

$g(\sin\theta - \cos\theta)$

B

$g(\sin\theta - \mu \cos\theta)$

C

$g(\mu \sin\theta - \cos\theta)$

D

$g\mu(\sin\theta - \cos\theta)$

Question 87: Palette 87/105

DPT MCQ

Concept

+

For a block held stationary against a vertical wall by an applied horizontal force $F$, the normal reaction force $N$ exerted by the wall equals the applied horizontal force ($N = F$). The maximum static friction force balancing the weight $W$ of the block is $f_{max} = \mu N$. When the force is just sufficient to prevent slipping, the weight of the block equals this maximum static frictional force.
Formula: $W = f_{max} = \mu N = \mu F$

A horizontal force of 10 N is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is 0.2. The weight of the block is

A

2 N

B

20 N

C

50 N

D

100 N

Question 88: Palette 88/105

DPT MCQ

Concept

+

When connected blocks move over pulleys, the net driving force overcomes the opposing forces such as friction and the weight of opposing blocks. According to Newton's Second Law, $F_{net} = m_{total} \cdot a$. The kinetic friction force is given by $f_k = \mu_k N = \mu_k m g$.

Find the acceleration of each block for the given system.

A

$1.5 \text{ ms}^{-2}$

B

$2.5 \text{ ms}^{-2}$

C

$0.5 \text{ ms}^{-2}$

D

$3.0 \text{ ms}^{-2}$

Question 89: Palette 89/105

DPT MCQ

Concept

+

For a block on an inclined plane connected via a string to a hanging block:
Normal force on the inclined plane: $N = m_1 g \cos\theta$
Component of weight along the incline: $F_{gravity} = m_1 g \sin\theta$
Maximum static friction force: $(f_s)_{max} = \mu_s N = \mu_s m_1 g \cos\theta$
Kinetic friction force: $f_k = \mu_k N = \mu_k m_1 g \cos\theta$
Net driving force along the direction of motion: $F_{net} = m_2 g - m_1 g \sin\theta - f_k$
Acceleration of the system: $a = \frac{F_{net}}{m_1 + m_2}$

Find acc. and friction.

A

$a = 1.0 \text{ ms}^{-2}$, $f = 32 \text{ N}$

B

$a = 10/11 \text{ ms}^{-2}$, $f = 32 \text{ N}$

C

$a = 10/21 \text{ ms}^{-2}$, $f = 40 \text{ N}$

D

$a = 2 \text{ ms}^{-2}$, $f = 40 \text{ N}$

Question 90: Palette 90/105

DPT MCQ

Concept

+

For a connected system consisting of a block on an inclined plane and a hanging mass connected by a string over a pulley, the normal force on the incline is $N = mg \cos\theta$. The maximum static friction force is $(f_s)_{max} = \mu_s N = \mu_s mg \cos\theta$ and the kinetic friction force is $f_k = \mu_k N = \mu_k mg \cos\theta$. According to Newton's Second Law, the acceleration of the combined system is given by $a = \frac{F_{net}}{m_{total}}$, where $F_{net}$ is the net pulling force minus opposing forces (downward component of weight and kinetic friction).

Find a?

A

$\frac{5}{21} \text{ ms}^{-2}$

B

$\frac{10}{21} \text{ ms}^{-2}$

C

$\frac{15}{21} \text{ ms}^{-2}$

D

$\frac{20}{21} \text{ ms}^{-2}$

Question 91: Palette 91/105

DPT MCQ

Concept

+

When a block is pressed against a vertical wall by a horizontal force $F$, the normal force from the wall is $N = F$.
The maximum static friction force is $(f_s)_{max} = \mu_s N$.
The kinetic friction force is $f_k = \mu_k N$.
The downward gravitational force is $mg$.
If $mg > (f_s)_{max}$, the block slides down with kinetic friction $f_k$ and acceleration $a = \frac{mg - f_k}{m}$.
If $mg \le (f_s)_{max}$, the block remains stationary, so $f = mg$ and $a = 0$.

Case I: Determine the magnitude of frictional force and acceleration of the block of mass 5 kg pushed against a vertical wall with a normal force of 100 N. [Take g = 10 m/s$^2$]

A

$f = 30 \text{ N}, a = 4 \text{ m/s}^2$

B

$f = 40 \text{ N}, a = 0 \text{ m/s}^2$

C

$f = 30 \text{ N}, a = 0 \text{ m/s}^2$

D

$f = 50 \text{ N}, a = 2 \text{ m/s}^2$

Question 92: Palette 92/105

DPT MCQ

Concept

+

When a block is pressed against a vertical wall by a horizontal force $F$, the normal force from the wall is $N = F$.
The maximum static friction force is $(f_s)_{max} = \mu_s N$.
If the downward force due to gravity $mg$ is less than or equal to $(f_s)_{max}$, static friction balances the weight ($f = mg$) and the block does not move ($a = 0$).

Case II: Determine the magnitude of frictional force and acceleration of the block of mass 5 kg pushed against a vertical wall with a normal force of 500 N. [Take g = 10 m/s$^2$]

A

$f = 200 \text{ N}, a = 4 \text{ m/s}^2$

B

$f = 50 \text{ N}, a = 0 \text{ m/s}^2$

C

$f = 150 \text{ N}, a = 0 \text{ m/s}^2$

D

$f = 50 \text{ N}, a = 2 \text{ m/s}^2$

Question 93: Palette 93/105

DPT MCQ

Concept

+

When an inclined pushing force $F$ is applied to a block against a vertical wall at an angle $\theta$ to the horizontal:
Horizontal component (Normal force): $N = F \cos\theta$.
Vertical upward component of pushing force: $F \sin\theta$.
Downward gravitational force: $mg$.
Net driving force along the vertical: $F_{driving} = F \sin\theta - mg$.
If $F_{driving} > (f_s)_{max} = \mu_s N$, the block moves upwards with kinetic friction $f_k = \mu_k N$ opposing the motion, and acceleration $a = \frac{F_{driving} - f_k}{m}$.

Case III: Determine the magnitude of frictional force and acceleration of the block of mass 12 kg pushed against a vertical wall by a force of 500 N inclined at an angle of 37$^\circ$ to the horizontal. [Take g = 10 m/s$^2$]

A

$f = 120 \text{ N}, a = 5 \text{ m/s}^2$

B

$f = 160 \text{ N}, a = 2 \text{ m/s}^2$

C

$f = 120 \text{ N}, a = 0 \text{ m/s}^2$

D

$f = 150 \text{ N}, a = 5 \text{ m/s}^2$

Question 94: Palette 94/105

DPT MCQ

Concept

+

When a block moves on a rough horizontal surface, kinetic friction opposes its motion and produces a constant deceleration $a = \mu g$. The distance traveled before coming to rest ($V_f = 0$) can be found using the third equation of motion: $v^2 = u^2 - 2ax$, where $u$ is the initial velocity and $x$ is the stopping distance.

In following fig. Find distance travel by block before coming to rest.

A

$30\text{ m}$

B

$40\text{ m}$

C

$50\text{ m}$

D

$60\text{ m}$

Question 95: Palette 95/105

DPT MCQ

Concept

+

When a block is projected up a rough inclined plane, both gravity and friction oppose its motion.
Deceleration along the incline: $a = g\sin\theta + \mu_k g\cos\theta$
Stopping distance: Using the kinematic relation $v^2 = u^2 - 2ax$, where $v = 0$, $u$ is initial velocity, and $x$ is distance.

In the given fig. block is projected along the rough incline ($\mu = 0.5$) with speed $50\text{ m/s}$. Find distance travel by block before coming to rest.

A

$100\text{ m}$

B

$125\text{ m}$

C

$150\text{ m}$

D

$175\text{ m}$

Question 96: Palette 96/105

DPT MCQ

Concept

+

For the $10\text{ kg}$ block to just start moving, the horizontal component of the tension force pulling it forward must equal the maximum static friction force opposing it.
Tension in the string: $T = mg$
Vertical equilibrium of the $10\text{ kg}$ block: $N + T\sin 37^\circ = M g \implies N = 100 - T\sin 37^\circ$
Maximum static friction force: $f_{max} = \mu N$
Horizontal equilibrium for impending motion: $T\cos 37^\circ = f_{max} = \mu (100 - T\sin 37^\circ)$

Find min value of m so that 10 kg block just move

A

$1.16\text{ kg}$

B

$2.32\text{ kg}$

C

$3.50\text{ kg}$

D

$0.85\text{ kg}$

Question 97: Palette 97/105

DPT MCQ

Concept

+

Two-Block Friction Problem:
To determine if two stacked blocks move together or separately, first assume they move together with a common acceleration $a = \frac{F_{\text{net}}}{m_{\text{total}}}$.
Calculate the required static friction force $f$ to accelerate the block that has no external force directly applied to it using Newton's Second Law: $F_{\text{net}} = m \cdot a$.
Compare the required friction $f$ with the maximum available static friction $f_{\text{max}} = \mu N = \mu m_{\text{A}} g$.
If $f \le f_{\text{max}}$, the blocks move together with common acceleration $a$. If $f > f_{\text{max}}$, they slide relative to each other and move separately under kinetic friction.

Find the acceleration of the two blocks. The system is initially at rest and the friction coefficient between block A and block B is $\mu = 0.5$. The ground below block B is smooth. A force $F = 50\text{ N}$ is applied horizontally to block A. Both blocks have a mass of $10\text{ kg}$.

A

$a_{\text{A}} = 5\text{ m/s}^2, a_{\text{B}} = 0\text{ m/s}^2$

B

$a_{\text{A}} = a_{\text{B}} = 2.5\text{ m/s}^2$

C

$a_{\text{A}} = 3\text{ m/s}^2, a_{\text{B}} = 2\text{ m/s}^2$

D

$a_{\text{A}} = a_{\text{B}} = 5\text{ m/s}^2$

Question 98: Palette 98/105

DPT MCQ

Concept

+

Two-Block Friction Problem:
1. Maximum Static Friction: $f_{\text{max}} = \mu N = \mu m_{\text{A}} g$.
2. Assume the blocks move together with common acceleration: $a = \frac{F}{m_{\text{A}} + m_{\text{B}}}$.
3. Calculate the static friction required to accelerate the bottom block B: $f_{\text{required}} = m_{\text{B}} \cdot a$.
4. Check relative motion:
* If $f_{\text{required}} \le f_{\text{max}}$, the blocks move together with common acceleration $a$.
* If $f_{\text{required}} > f_{\text{max}}$, slipping occurs, and kinetic friction $f_{\text{k}} = f_{\text{max}}$ acts between the blocks. In this case, calculate accelerations separately: $a_{\text{A}} = \frac{F - f_{\text{k}}}{m_{\text{A}}}$ and $a_{\text{B}} = \frac{f_{\text{k}}}{m_{\text{B}}}$.

Find the acceleration of the two blocks. The system is initially at rest and the friction coefficient are as shown in the figure?

A

$a_{\text{A}} = 5.05\text{ m/s}^2, a_{\text{B}} = 5.05\text{ m/s}^2$

B

$a_{\text{A}} = 5.1\text{ m/s}^2, a_{\text{B}} = 5\text{ m/s}^2$

C

$a_{\text{A}} = 5\text{ m/s}^2, a_{\text{B}} = 5.1\text{ m/s}^2$

D

$a_{\text{A}} = 10.1\text{ m/s}^2, a_{\text{B}} = 0\text{ m/s}^2$

Question 99: Palette 99/105

DPT MCQ

Concept

+

Two-Block Friction Problem:
1. Maximum Available Static Friction: $f_{\text{max}} = \mu N = \mu m_{\text{top}} g$.
2. Assume the blocks move together with a common acceleration: $a = \frac{F}{m_{\text{top}} + m_{\text{bottom}}}$.
3. Calculate the static friction required to accelerate the lower block: $f_{\text{required}} = m_{\text{bottom}} \cdot a$.
4. Check relative motion:
* If $f_{\text{required}} \le f_{\text{max}}$, the blocks move together with common acceleration $a$.
* If $f_{\text{required}} > f_{\text{max}}$, slipping occurs between the blocks and they move separately.

Find the acceleration of the two blocks. The system is initially at rest and the friction coefficient are as shown in the figure?

A

$a_{\text{A}} = 6\text{ m/s}^2, a_{\text{B}} = 0\text{ m/s}^2$

B

$a_{\text{A}} = a_{\text{B}} = 2\text{ m/s}^2$

C

$a_{\text{A}} = 3\text{ m/s}^2, a_{\text{B}} = 2\text{ m/s}^2$

D

$a_{\text{A}} = a_{\text{B}} = 3\text{ m/s}^2$

Question 100: Palette 100/105

DPT MCQ

Concept

+

Minimum Force for Relative Sliding:
Sliding between two blocks starts when the friction force between them reaches its maximum limiting value, $f_{\text{max}} = \mu N = \mu m_{\text{top}} g$.
At the verge of sliding (just sliding condition), both blocks still move together with a common acceleration $a$, but the static friction between them is maximum ($f = f_{\text{max}}$).
For the top block driven purely by static friction: $f_{\text{max}} = m_{\text{top}} \cdot a \implies a = \frac{f_{\text{max}}}{m_{\text{top}}}$.
For the bottom block driven by external force $F$: $F - f_{\text{max}} = m_{\text{bottom}} \cdot a$.
Substituting acceleration $a$ gives the minimum force $F_{\text{min}}$ required to initiate sliding between the blocks.

Initially the system is at rest. find out minimum value of F for which sliding starts between the two blocks.

A

$100\text{ N}$

B

$120\text{ N}$

C

$150\text{ N}$

D

$200\text{ N}$

Question 101: Palette 101/105

DPT MCQ

Concept

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Two-Block Friction System:
1. Friction between contact surfaces depends on the coefficient of friction ($\mu$): $f_{\text{max}} = \mu N$.
2. Friction force between Block A and Block B: Since $\mu = 0$, no friction acts between A and B ($f_{\text{AB}} = 0$). Block A accelerates independently under the applied force $F$: $a_{\text{A}} = \frac{F}{m_{\text{A}}}$.
3. Friction force between Block B and Ground G: Since no horizontal force is transferred to block B (as $f_{\text{AB}} = 0$), block B remains stationary ($a_{\text{B}} = 0$), and static friction between B and ground is zero ($f_{\text{BG}} = 0$).

In the given figure, find the accelerations and the friction forces involved for block A ($5\text{ kg}$) and block B ($10\text{ kg}$) when a force $F = 15\text{ N}$ is applied to block A. The friction coefficient between A and B is $\mu = 0$, and between B and Ground G is $\mu' = 0.5$.

A

$a_{\text{A}} = 3\text{ m/s}^2, a_{\text{B}} = 0, f_{\text{AB}} = 0, f_{\text{BG}} = 0$

B

$a_{\text{A}} = 1\text{ m/s}^2, a_{\text{B}} = 1\text{ m/s}^2, f_{\text{AB}} = 5\text{ N}, f_{\text{BG}} = 10\text{ N}$

C

$a_{\text{A}} = 3\text{ m/s}^2, a_{\text{B}} = 1.5\text{ m/s}^2, f_{\text{AB}} = 0, f_{\text{BG}} = 25\text{ N}$

D

$a_{\text{A}} = 0, a_{\text{B}} = 0, f_{\text{AB}} = 15\text{ N}, f_{\text{BG}} = 15\text{ N}$

Question 102: Palette 102/105

DPT MCQ

Concept

+

Two-Block Friction System:
1. Maximum static friction between block A and block B: $f_{\text{AB, max}} = \mu m_{\text{A}} g$.
2. Maximum static friction between block B and ground G: $f_{\text{BG, max}} = \mu' (m_{\text{A}} + m_{\text{B}}) g$.
3. Net external force on block A tries to slide it over block B. The friction force $f_{\text{AB}}$ acts backward on block A and forward on block B.
4. For block B to move, the friction force transferred to it from block A ($f_{\text{AB}}$) must exceed the maximum static friction at the ground surface ($f_{\text{BG, max}}$).
5. If $f_{\text{AB}} \le f_{\text{BG, max}}$, block B remains stationary ($a_{\text{B}} = 0$) and ground friction balances $f_{\text{AB}}$.

In the given figure, find the accelerations and the friction forces involved for block A ($5\text{ kg}$) and block B ($10\text{ kg}$) when a force $F = 30\text{ N}$ is applied to block A. The friction coefficient between A and B is $\mu = 0.5$, and between B and Ground G is $\mu' = 0.5$.

A

$a_{\text{A}} = 1\text{ m/s}^2, a_{\text{B}} = 0, f_{\text{AB}} = 25\text{ N}, f_{\text{BG}} = 25\text{ N}$

B

$a_{\text{A}} = 2\text{ m/s}^2, a_{\text{B}} = 1\text{ m/s}^2, f_{\text{AB}} = 20\text{ N}, f_{\text{BG}} = 25\text{ N}$

C

$a_{\text{A}} = 1\text{ m/s}^2, a_{\text{B}} = 1\text{ m/s}^2, f_{\text{AB}} = 25\text{ N}, f_{\text{BG}} = 25\text{ N}$

D

$a_{\text{A}} = 3\text{ m/s}^2, a_{\text{B}} = 0, f_{\text{AB}} = 15\text{ N}, f_{\text{BG}} = 15\text{ N}$

Question 103: Palette 103/105

DPT MCQ

Concept

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Two-Block Friction System:
1. Maximum friction between block A and block B: $f_{\text{AB, max}} = \mu m_{\text{A}} g$.
2. Maximum friction between block B and Ground G: $f_{\text{BG, max}} = \mu' (m_{\text{A}} + m_{\text{B}}) g$.
3. Block A is accelerated solely by the friction force $f_{\text{AB}}$ exerted by block B. Therefore, the maximum possible acceleration of block A before slipping occurs is $a_{\text{A, max}} = \frac{f_{\text{AB, max}}}{m_{\text{A}}}$.
4. If the acceleration required to move both blocks together exceeds $a_{\text{A, max}}$, slipping occurs between A and B. In this case, kinetic friction $f_{\text{AB}} = f_{\text{AB, max}}$ acts, and both blocks move separately.

In the given figure, find the accelerations and the friction forces involved for block A ($5\text{ kg}$) and block B ($10\text{ kg}$) when a horizontal force $F = 200\text{ N}$ is applied to block B. The friction coefficient between A and B is $\mu = 0.5$, and between B and Ground G is $\mu' = 0.5$.

A

$a_{\text{A}} = 1\text{ m/s}^2, a_{\text{B}} = 1\text{ m/s}^2, f_{\text{AB}} = 5\text{ N}, f_{\text{BG}} = 75\text{ N}$

B

$a_{\text{A}} = 5\text{ m/s}^2, a_{\text{B}} = 10\text{ m/s}^2, f_{\text{AB}} = 25\text{ N}, f_{\text{BG}} = 75\text{ N}$

C

$a_{\text{A}} = 3\text{ m/s}^2, a_{\text{B}} = 0, f_{\text{AB}} = 0, f_{\text{BG}} = 0$

D

$a_{\text{A}} = 1\text{ m/s}^2, a_{\text{B}} = 0, f_{\text{AB}} = 25\text{ N}, f_{\text{BG}} = 25\text{ N}$

Question 104: Palette 104/105

DPT MCQ

Concept

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Two-Block Friction System:
1. Maximum static friction between block A and block B: $f_{\text{AB, max}} = \mu m_{\text{A}} g$.
2. Maximum static friction between block B and Ground G: $f_{\text{BG, max}} = \mu' (m_{\text{A}} + m_{\text{B}}) g$.
3. Assume both blocks move together with a common acceleration: $a = \frac{F - f_{\text{BG, max}}}{m_{\text{A}} + m_{\text{B}}}$.
4. Determine the required static friction force to accelerate block A at this rate: $f_{\text{AB, req}} = m_{\text{A}} \cdot a$.
5. If $f_{\text{AB, req}} \le f_{\text{AB, max}}$, the blocks move together with common acceleration $a_{\text{A}} = a_{\text{B}} = a$, and the ground friction is $f_{\text{BG}} = f_{\text{BG, max}}$.

In the given figure, find the accelerations and the friction forces involved for block A ($5\text{ kg}$) and block B ($10\text{ kg}$) when a force $F = 90\text{ N}$ is applied to block B. The friction coefficient between A and B is $\mu = 0.5$, and between B and Ground G is $\mu' = 0.5$.

A

$a_{\text{A}} = a_{\text{B}} = 1\text{ m/s}^2, f_{\text{AB}} = 5\text{ N}, f_{\text{BG}} = 75\text{ N}$

B

$a_{\text{A}} = 5\text{ m/s}^2, a_{\text{B}} = 10\text{ m/s}^2, f_{\text{AB}} = 25\text{ N}, f_{\text{BG}} = 75\text{ N}$

C

$a_{\text{A}} = 3\text{ m/s}^2, a_{\text{B}} = 0, f_{\text{AB}} = 0, f_{\text{BG}} = 0$

D

$a_{\text{A}} = 1\text{ m/s}^2, a_{\text{B}} = 0, f_{\text{AB}} = 25\text{ N}, f_{\text{BG}} = 25\text{ N}$

Question 105: Palette 105/105

DPT MCQ

Concept

+

Two-Block System without Relative Motion:
1. Since there is no relative motion, both blocks move together with a common acceleration $a = \frac{F_{\text{net, external}}}{m_1 + m_2}$.
2. The net horizontal force on the top block ($2\text{ kg}$) satisfies Newton's Second Law: $F_{\text{net, top}} = m_1 \cdot a = 2 \cdot a$.
3. The friction force $f$ on the $2\text{ kg}$ block is calculated by applying $\sum F_x = m_1 \cdot a$ on it.
4. Direct calculation for each figure in Column II:
* For (p): $F_{\text{net}} = 12\text{ N} \implies a = 2\text{ m/s}^2 \implies f = 4\text{ N}$ (towards right).
* For (q): $F_{\text{net}} = 12\text{ N} \implies a = 2\text{ m/s}^2 \implies 12 - f = 2(2) \implies f = 8\text{ N}$ (towards left).
* For (r): $F_{\text{net}} = 12\text{ N} \implies a = 2\text{ m/s}^2 \implies 6 - f = 2(2) \implies f = 2\text{ N}$ (towards left).
* For (s): $F_{\text{net}} = 20 - 8 = 12\text{ N} \implies a = 2\text{ m/s}^2 \implies f - 8 = 2(2) \implies f = 12\text{ N}$ (towards right).

Column II gives certain situations involving two blocks of mass $2\text{ kg}$ and $4\text{ kg}$. The $4\text{ kg}$ block lies on a smooth horizontal table. There is sufficient friction between both the blocks and there is no relative motion between the blocks in all situations. Horizontal forces act on one or both blocks as shown. Column I gives certain statements related to figures given in column II. Match the statements in column I with the figure in column II.

A

(A) - s ; (B) - r ; (C) - p, s ; (D) - q, r

B

(A) - r ; (B) - q ; (C) - p ; (D) - s

C

(A) - p ; (B) - s ; (C) - q ; (D) - r

D

(A) - q ; (B) - p ; (C) - r ; (D) - s