Mix Examples-Newton's Laws of Motion and Friction Questions in English

(C) The block remains at rest as long as the applied force $F$ is less than or equal to the maximum static friction force $f_{s,max} = \mu mg$.
So,the block starts sliding when $F = \mu mg$,which implies $Kt = \mu mg$,or $t = \frac{\mu mg}{K}$.
For $t \leq \frac{\mu mg}{K}$,the acceleration $a = 0$.
For $t > \frac{\mu mg}{K}$,the net force acting on the block is $F_{net} = F - f_k = Kt - \mu mg$,where $f_k = \mu mg$ is the kinetic friction.
The acceleration is given by $a = \frac{F_{net}}{m} = \frac{Kt - \mu mg}{m} = \frac{K}{m}t - \mu g$.
This equation represents a straight line with a positive slope $\frac{K}{m}$ and a negative intercept $-\mu g$ on the acceleration axis.
Thus,the acceleration is zero until $t = \frac{\mu mg}{K}$ and then increases linearly with time. This corresponds to the graph shown in option $C$.

(C) The block of weight $60 \, N$ is on a rough surface with coefficient of friction $\mu = 0.5$. The maximum frictional force $F_{max}$ that can act on the block is given by $F_{max} = \mu N = 0.5 \times 60 \, N = 30 \, N$.
For the block not to slip,the tension $T_1$ in the horizontal string must be less than or equal to $F_{max}$. Thus,$T_1 = 30 \, N$.
At the junction point,the forces are in equilibrium. Let $T_2$ be the tension in the inclined string at an angle of $45^{\circ}$ to the horizontal.
The horizontal component of $T_2$ balances $T_1$: $T_2 \cos 45^{\circ} = T_1 = 30 \, N$.
$T_2 \times \frac{1}{\sqrt{2}} = 30 \implies T_2 = 30\sqrt{2} \, N$.
The vertical component of $T_2$ balances the weight $W$: $W = T_2 \sin 45^{\circ}$.
$W = 30\sqrt{2} \times \frac{1}{\sqrt{2}} = 30 \, N$.
Therefore,the maximum weight $W$ is $30 \, N$.

(A) Let $F$ be the upward buoyant force acting on the balloon.
When the balloon is descending with acceleration $a = \frac{g}{2}$,the equation of motion is:
$mg - F = m \left(\frac{g}{2}\right)$
$F = mg - \frac{mg}{2} = \frac{mg}{2} \quad ...(1)$
Let $m_1$ be the mass removed. The new mass of the balloon is $(m - m_1)$.
When the balloon moves up with acceleration $a = \frac{g}{2}$,the equation of motion is:
$F - (m - m_1)g = (m - m_1) \left(\frac{g}{2}\right)$
$F = (m - m_1) \left(g + \frac{g}{2}\right) = \frac{3}{2}(m - m_1)g \quad ...(2)$
Equating $(1)$ and $(2)$:
$\frac{mg}{2} = \frac{3}{2}(m - m_1)g$
$m = 3(m - m_1)$
$m = 3m - 3m_1$
$3m_1 = 2m$
$m_1 = \frac{2m}{3}$

(B) The force exerted by a fluid jet on a wall is given by the rate of change of momentum.
$F = \frac{dp}{dt} = \frac{d(mv)}{dt} = v \frac{dm}{dt}$.
Since the mass flow rate $\frac{dm}{dt} = \rho A v$,the force is $F = \rho A v^2$.
Given:
Density $\rho = 1000\, kg/m^3$
Area $A = 10^{-3}\, m^2$
Velocity $v = 20\, m/s$
Substituting the values:
$F = 1000 \times 10^{-3} \times (20)^2$
$F = 1 \times 400 = 400\, N$.

(B) The system consists of two blocks of masses $m_1 = 4\, kg$ and $m_2 = 2\, kg$ sliding down an inclined plane with angle $\theta = 30^{\circ}$ and coefficient of kinetic friction $\mu = 0.3$.
Since the blocks are in contact and moving together,they will have the same acceleration $a$.
The net driving force down the plane is the component of gravity: $F_g = (m_1 + m_2)g \sin \theta$.
The total frictional force opposing the motion is: $f_k = \mu(m_1 + m_2)g \cos \theta$.
Applying Newton's second law for the system: $(m_1 + m_2)a = (m_1 + m_2)g \sin \theta - \mu(m_1 + m_2)g \cos \theta$.
Dividing by $(m_1 + m_2)$,we get: $a = g(\sin \theta - \mu \cos \theta)$.
Substituting the values ($g = 9.8\, m/s^2$,$\sin 30^{\circ} = 0.5$,$\cos 30^{\circ} \approx 0.866$):
$a = 9.8 \times (0.5 - 0.3 \times 0.866) = 9.8 \times (0.5 - 0.2598) = 9.8 \times 0.2402 \approx 2.354\, m/s^2$.
Rounding to the nearest option,the acceleration is approximately $2.4\, m/s^2$.

(B) The box is moving from left to right with velocity $V$. When the train decelerates,the box tends to continue its motion due to inertia,so it experiences a pseudo force towards the left relative to the train.
Consequently,a kinetic friction force $f_r$ acts on the box towards the left to oppose this relative motion.
The floor also exerts a normal force $N$ acting vertically upwards on the box.
The net reaction force $R$ exerted by the floor on the box is the vector sum of the normal force $N$ and the friction force $f_r$,i.e.,$R = N + f_r$.
Since $N$ is upwards and $f_r$ is towards the left,the resultant reaction force $R$ will point upwards and to the left.
This corresponds to the direction shown in option $B$.

(A) Friction arises due to the intermolecular forces between the surfaces in contact. These intermolecular forces are essentially electromagnetic in nature,as they result from the interactions between the charged particles (electrons and nuclei) of the atoms on the two surfaces. Therefore,friction is classified as an electromagnetic force.

(C) The block remains at rest as long as the applied force $F = kt$ is less than or equal to the maximum static friction $f_{max} = \mu mg$.
Thus, the block starts moving when $kt \geq \mu mg$, which means $t \geq \frac{\mu mg}{k} = t_0$.
For $t < t_0$, the velocity $v = 0$.
For $t \geq t_0$, the net force on the block is $F_{net} = kt - \mu mg$.
Using Newton's second law, $ma = kt - \mu mg$, so the acceleration is $a = \frac{k}{m}t - \mu g$.
Since $a = \frac{dv}{dt}$, we integrate with respect to time: $v = \int_{t_0}^{t} (\frac{k}{m}t' - \mu g) dt'$.
This results in $v = \frac{k}{2m}(t^2 - t_0^2) - \mu g(t - t_0)$.
Since the velocity is proportional to $t^2$, the graph is parabolic for $t > t_0$.

(B) The force required to accelerate the car is given by Newton's second law: $F_{net} = ma$.
Here,$F_{net} = F_{engine} - F_{resistance}$.
Given $m = 500\, kg$,$a = 1\, m/s^2$,and $F_{resistance} = 1000\, N$.
$F_{engine} - 1000 = 500 \times 1$.
$F_{engine} = 1500\, N$.
The power $P$ delivered by the engine is given by $P = F_{engine} \times v$.
Given $v = 5\, m/s$.
$P = 1500 \times 5 = 7500\, W$.
Converting to kilowatts: $P = 7.5\, kW$.

(C) Let the total force exerted by the five persons be $\vec{F}_{total} = \vec{F}_A + \vec{F}_B + \vec{F}_C + \vec{F}_D + \vec{F}_E$. Given $m = 100 \, kg$ and $\vec{a} = 3 \hat{i} \, m/s^2$ (taking East as $+\hat{i}$),we have $\vec{F}_{total} = 100 \times 3 \hat{i} = 300 \hat{i} \, N$ $...(1)$.
When $A$ stops,the force is $\vec{F}_{total} - \vec{F}_A = 100 \times (-1 \hat{i}) = -100 \hat{i} \, N$ $...(2)$.
Subtracting $(2)$ from $(1)$,we get $\vec{F}_A = 400 \hat{i} \, N$.
When $B$ stops,the force is $\vec{F}_{total} - \vec{F}_B = 100 \times (24 \hat{j}) = 2400 \hat{j} \, N$ (taking North as $+\hat{j}$) $...(3)$.
Subtracting $(3)$ from $(1)$,we get $\vec{F}_B = 300 \hat{i} - 2400 \hat{j} \, N$.
When only $A$ and $B$ pull,the net force is $\vec{F}_{net} = \vec{F}_A + \vec{F}_B = 400 \hat{i} + 300 \hat{i} - 2400 \hat{j} = 700 \hat{i} - 2400 \hat{j} \, N$.
The acceleration is $\vec{a} = \frac{\vec{F}_{net}}{m} = \frac{700 \hat{i} - 2400 \hat{j}}{100} = 7 \hat{i} - 24 \hat{j} \, m/s^2$.
The magnitude is $|\vec{a}| = \sqrt{7^2 + (-24)^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \, m/s^2$.

(D) The normal force $N$ exerted by the wall on the block is equal to the applied horizontal force: $N = 12 \, N$.
The maximum static frictional force $f_{s,max}$ is given by $f_{s,max} = \mu_s N = 0.5 \times 12 \, N = 6 \, N$.
The weight of the block is $W = mg = (1/2) \, kg \times 10 \, m/s^2 = 5 \, N$.
Since the weight $W = 5 \, N$ is less than the maximum static friction $f_{s,max} = 6 \, N$,the block will not move.
The wall exerts two forces on the block: the normal force $N = 12 \, N$ (horizontal) and the static frictional force $f_s = 5 \, N$ (vertical,balancing the weight).
The total force exerted by the wall on the block is the resultant of these two perpendicular forces: $F_{total} = \sqrt{N^2 + f_s^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \, N$.

(C) For the block,the forces are balanced as follows:
$T \cos 37^{\circ} = f$
$N + T \sin 37^{\circ} = Mg$
Since $f = \mu N$,we have $T \cos 37^{\circ} = \mu (Mg - T \sin 37^{\circ})$.
Substituting the values $M = 100 \, kg$,$\mu = 1/3$,$\cos 37^{\circ} = 4/5$,and $\sin 37^{\circ} = 3/5$:
$T(4/5) = (1/3)(100g - T(3/5))$
$12T/5 = 100g - 3T/5$
$15T/5 = 100g \implies 3T = 100g \implies T = 100g/3$.
For the person,the upward force is $T = 100g/3$ and the downward force is $mg = 25g$.
The equation of motion is $T - mg = ma_{\max}$.
$100g/3 - 25g = 25a_{\max}$
$(100g - 75g)/3 = 25a_{\max}$
$25g/3 = 25a_{\max}$
$a_{\max} = g/3$.

(C) $1$. Initially,for block $A$,the applied force $F=t$ is balanced by static friction $f_{rA}$. The block $A$ remains at rest until $F = \mu_s mg$,i.e.,$t = \mu_s mg$. During this time,the tension $T$ in the string is $0$,so block $B$ remains at rest and the friction force on $B$ is $0$.
$2$. When $t > \mu_s mg$,block $A$ starts to move. The tension $T$ in the string increases as $T = F - f_{rA} = t - \mu_k mg$ (assuming kinetic friction $\mu_k mg$ acts on $A$).
$3$. Block $B$ will start to move when the tension $T$ exceeds the maximum static friction force on $B$,which is $\mu_s mg$. So,$T = t - \mu_k mg = \mu_s mg$,which gives $t = \mu_s mg + \mu_k mg$.
$4$. For $t < \mu_s mg + \mu_k mg$,block $B$ is at rest,so the friction force on $B$ is equal to the tension $T = t - \mu_k mg$. However,since $T$ is only non-zero for $t > \mu_s mg$,the friction on $B$ is $0$ until $t = \mu_s mg$,then it increases linearly with $t$.
$5$. Once $t$ reaches the threshold where $B$ moves,the friction force on $B$ drops from its maximum static value $\mu_s mg$ to the kinetic friction value $\mu_k mg$ and remains constant.
$6$. Comparing this behavior with the given options,graph $C$ correctly represents this: zero friction until $t = \mu_s mg$,then increasing friction,a peak at the threshold,and finally a constant kinetic friction value.

(C) For the $5 \, kg$ block,the maximum static friction force is $f_{s1} = \mu m_1 g = 0.2 \times 5 \times 10 = 10 \, N$.
For the $10 \, kg$ block,the maximum static friction force is $f_{s2} = \mu m_2 g = 0.2 \times 10 \times 10 = 20 \, N$.
For the system to remain at rest,the applied force $F$ must be balanced by the total maximum static friction of the system.
However,the $5 \, kg$ block will only move if the force transmitted to it exceeds $10 \, N$. Since the force $F$ is applied to the $10 \, kg$ block,the entire system will start moving when $F$ exceeds the total friction $f_{s1} + f_{s2} = 10 + 20 = 30 \, N$.
Thus,the maximum force $F$ that will not cause motion of any block is $30 \, N$.

(C) At the bottom of the bowl,the normal force $N$ and gravity $mg$ act on the particle. The net force provides the centripetal acceleration:
$N - mg = \frac{mv_0^2}{R}$
Given $v_0 = \sqrt{gR}$,we have:
$N - mg = \frac{m(gR)}{R} = mg$
$N = 2mg$
The kinetic friction force is $f_k = \mu N = 0.5 \times 2mg = mg$.
The tangential acceleration $a_T$ is caused by friction:
$ma_T = f_k = mg \implies a_T = g$ (directed horizontally,opposite to velocity).
The normal acceleration $a_N$ is the centripetal acceleration:
$a_N = \frac{v_0^2}{R} = \frac{gR}{R} = g$ (directed vertically upwards).
The total acceleration $a$ is the vector sum of $a_T$ and $a_N$:
$a = \sqrt{a_T^2 + a_N^2} = \sqrt{g^2 + g^2} = \sqrt{2}g$.
The direction is at an angle of $45^\circ$ to the horizontal,pointing upwards and backwards,represented as $\nwarrow$.

(A) For $t \le 1\,sec$,the truck accelerates at $a_T = 5\,m/s^2$. The maximum acceleration the block can have due to friction is $a_{max} = \mu g = 0.2 \times 10 = 2\,m/s^2$. Since $a_T > a_{max}$,the block slips. The block accelerates at $2\,m/s^2$ for the entire duration until its velocity matches the truck's velocity.
At $t = 1\,sec$,the truck's velocity is $v_T = a_T \times t = 5 \times 1 = 5\,m/s$. The block's velocity is $v_B = a_{max} \times t = 2 \times 1 = 2\,m/s$.
After $t = 1\,sec$,the truck moves at a constant velocity of $5\,m/s$. The block continues to accelerate at $2\,m/s^2$ until its velocity reaches $5\,m/s$.
Let $t'$ be the time taken for the block to reach $5\,m/s$ from $t = 1\,sec$. Using $v = u + at$,we have $5 = 2 + 2 \times t'$,which gives $t' = 1.5\,sec$.
The total time taken is $t_{total} = 1 + 1.5 = 2.5\,sec$. Thus,the velocity of the block increases linearly to $5\,m/s$ at $t = 2.5\,sec$ and remains constant thereafter.

(D) The forces acting on a body of mass $m$ are its weight $mg$ acting vertically downward and air resistance $F$ acting vertically upward.
The net force is $F_{net} = mg - F$.
The acceleration of the body is $a = \frac{F_{net}}{m} = g - \frac{F}{m}$.
For the body of mass $M$,the acceleration is $a_M = g - \frac{F}{M}$.
For the body of mass $m$,the acceleration is $a_m = g - \frac{F}{m}$.
Since $M > m$,it follows that $\frac{F}{M} < \frac{F}{m}$,which implies $a_M > a_m$.
The body with the larger mass $M$ will have a greater acceleration and will reach the ground first.
Therefore,the $Assertion$ is incorrect because they do not reach simultaneously,and the $Reason$ is also incorrect because the accelerations are not the same.

(A) To move a load of mass $m$ with a constant velocity on an inclined plane of angle $\theta$,the required force $F$ is $mg \sin \theta$.
In the given $F-x$ graph,the force is positive for the first half of the distance $L$ and negative for the second half.
This indicates that for the first half,the load is being pushed up an incline (requiring a positive force $mg \sin \theta$),and for the second half,the load is moving down an incline (where the force required to maintain constant velocity is $-mg \sin \theta$).
The surface profile that consists of an upward incline followed by a downward incline is the triangular profile shown in the first option image (labeled as option $A$ in the context of the provided images).

(A) The force exerted by the water on the gardener is given by the rate of change of momentum,$F = \frac{dp}{dt} = v \frac{dm}{dt}$.
Initially,the force is $F_1 = \frac{dm}{dt} \cdot v_1 = 4 \times 2 = 8\, N$.
When the speed increases to $3\, ms^{-1}$,the new force is $F_2 = \frac{dm}{dt} \cdot v_2 = 4 \times 3 = 12\, N$.
The change in force (jerk) experienced by the gardener is $\Delta F = F_2 - F_1 = 12 - 8 = 4\, N$.
Since the water is pushing the hosepipe backward,the increase in momentum requires an additional backward force,so the gardener experiences a jerk of $4\, N$ in the backward direction.

(B) For figure $(a)$:
Let $a$ be the acceleration of the system. The equations of motion are:
For mass $m$: $T - mg = ma$ $(i)$
For mass $2m$: $2mg - T = 2ma$ $(ii)$
Adding $(i)$ and $(ii)$,we get:
$mg = 3ma$
$\therefore a = g/3$
For figure $(b)$:
Let $a'$ be the acceleration of mass $m$. The tension in the rope is $T' = F = 2mg$ because the force is applied directly to the rope.
The equation of motion for mass $m$ is:
$T' - mg = ma'$
Substituting $T' = 2mg$:
$2mg - mg = ma'$
$mg = ma'$
$\therefore a' = g$
Thus,the accelerations are $g/3$ and $g$ respectively.

(C) The change in momentum occurs only in the direction perpendicular to the wall.
Let the velocity be $v = 10 \, m/s$ and mass $m = 3 \, kg$.
The component of velocity perpendicular to the wall is $v_{\perp} = v \sin(60^\circ)$.
Initial momentum perpendicular to the wall: $p_i = mv \sin(60^\circ)$.
Final momentum perpendicular to the wall (after reflection): $p_f = -mv \sin(60^\circ)$.
Change in momentum: $\Delta p = p_f - p_i = -mv \sin(60^\circ) - mv \sin(60^\circ) = -2mv \sin(60^\circ)$.
The magnitude of the change in momentum is $|\Delta p| = 2mv \sin(60^\circ)$.
Force exerted on the wall $F = \frac{|\Delta p|}{\Delta t} = \frac{2mv \sin(60^\circ)}{\Delta t}$.
Substituting the values: $F = \frac{2 \times 3 \times 10 \times \sin(60^\circ)}{0.2} = \frac{60 \times (\sqrt{3}/2)}{0.2} = \frac{30\sqrt{3}}{0.2} = 150\sqrt{3} \, N$.

(A) Let $a$ be the acceleration of the system. Since the string is inextensible,both blocks have the same magnitude of acceleration $a$.
For block $m_1$ on the table,the forces acting horizontally are the tension $T$ (towards the right) and the applied force $F = m_2g/2$ (towards the left). Assuming the block $m_1$ moves towards the right,the equation of motion is:
$T - F = m_1a$
$T - m_2g/2 = m_1a$ --- $(i)$
For block $m_2$ hanging vertically,the forces are gravity $m_2g$ (downwards) and tension $T$ (upwards). Since $m_1$ moves right,$m_2$ moves downwards:
$m_2g - T = m_2a$ --- $(ii)$
Adding equations $(i)$ and $(ii)$:
$(T - m_2g/2) + (m_2g - T) = m_1a + m_2a$
$m_2g/2 = (m_1 + m_2)a$
$a = \frac{m_2g}{2(m_1 + m_2)}$
Since the result is positive,the assumption that $m_1$ moves towards the right is correct.

(D) The initial velocity of the player is $\vec{v}_i = v\hat{j}$ (northward).
The final velocity of the player is $\vec{v}_f = -v\hat{i}$ (westward).
The change in velocity is $\Delta\vec{v} = \vec{v}_f - \vec{v}_i = -v\hat{i} - v\hat{j}$.
According to Newton's second law,the force $\vec{F}$ acting on the player is in the direction of the change in velocity $\Delta\vec{v}$.
The direction of $\Delta\vec{v} = -v\hat{i} - v\hat{j}$ is south-west.
Since the player changes their direction by applying force through their feet against the ground,this is a muscular force exerted by the player on the ground,and by Newton's third law,the ground exerts an equal and opposite force on the player. Thus,the force acting on the player is a muscular force directed towards the south-west.

(A) On a rainy day,the roads become wet.
Water acts as a lubricant between the tires and the road surface,which significantly lowers the coefficient of friction $(\mu)$.
Since the frictional force $f = \mu N$ (where $N$ is the normal force),a decrease in $\mu$ leads to a decrease in the available frictional force.
This reduced friction makes it difficult to maintain control,increases the braking distance,and increases the chances of skidding at high speeds.
Therefore,both the $Assertion$ and $Reason$ are correct,and the $Reason$ is the correct explanation for the $Assertion$.

(C) The $Assertion$ is correct because ball bearings are widely used in machines to reduce friction between moving parts.
The $Reason$ is incorrect because the primary purpose of using ball bearings is to convert sliding friction into rolling friction,which is significantly smaller than sliding friction. While they may provide stability,they are not primarily used to reduce vibrations in the context of friction reduction.

(D) The $Assertion$ states that the man cannot move on the horizontal surface while holding a rope connected to a block. This is incorrect because the man can exert a force on the block by pulling the rope. According to Newton's third law,the block will exert an equal and opposite force on the man,causing him to move towards the block.
The $Reason$ states that a man standing at rest on a smooth horizontal surface cannot start walking due to the absence of friction. This is correct. Walking requires friction to push the ground backward,which in turn provides a forward reaction force. Without friction,the man's feet would simply slip.
Since the $Assertion$ is incorrect and the $Reason$ is correct,the correct option is $D$ (based on standard assertion-reason logic where the assertion is false).

(A) Since the string is inextensible and the pulley is smooth,the $3\; kg$ block and the $20\; kg$ trolley have the same magnitude of acceleration $a$.
Applying Newton's second law to the motion of the $3\; kg$ block:
$30 - T = 3a$ --- (Equation $1$)
Applying Newton's second law to the motion of the $20\; kg$ trolley:
$T - f_k = 20a$ --- (Equation $2$)
Where the kinetic friction $f_k = \mu_k N$.
Given $\mu_k = 0.04$ and the normal force $N = m_{trolley} g = 20 \times 10 = 200\; N$.
So,$f_k = 0.04 \times 200 = 8\; N$.
Substituting $f_k$ into Equation $2$:
$T - 8 = 20a$ --- (Equation $3$)
Adding Equation $1$ and Equation $3$:
$(30 - T) + (T - 8) = 3a + 20a$
$22 = 23a$
$a = \frac{22}{23} \approx 0.96\; m s^{-2}$.
Substituting $a$ into Equation $1$:
$30 - T = 3 \times (0.9565)$
$T = 30 - 2.87 = 27.13\; N \approx 27.1\; N$.

(N/A) The block is at rest on the floor. Its free-body diagram shows two forces on the block: the force of gravitational attraction by the earth equal to $2 \times 10 = 20\; N$,and the normal force $R$ of the floor on the block. By the First Law,the net force on the block must be zero,i.e.,$R = 20\; N$. Using the third law,the action of the block (i.e.,the force exerted on the floor by the block) is equal to $20\; N$ and directed vertically downwards.
$(b)$ The system (block $+$ cylinder) accelerates downwards with $0.1\; m/s^2$. The free-body diagram of the system shows two forces on the system: the force of gravity due to the earth $(27\; kg \times 10\; m/s^2 = 270\; N)$; and the normal force $R'$ by the floor. Note that the free-body diagram of the system does not show the internal forces between the block and the cylinder. Applying the second law to the system:
$270 - R' = 27 \times 0.1\; N$
$R' = 270 - 2.7 = 267.3\; N$
By the third law,the action of the system on the floor is equal to $267.3\; N$ vertically downward.
Action-reaction pairs:
For $(a)$: $(i)$ The force of gravity $(20\; N)$ on the block by the earth (action); the force of gravity on the earth by the block (reaction) equal to $20\; N$ directed upwards. $(ii)$ The force on the floor by the block (action); the force on the block by the floor (reaction).
For $(b)$: $(i)$ The force of gravity $(270\; N)$ on the system by the earth (action); the force of gravity on the earth by the system (reaction),equal to $270\; N$ directed upwards. $(ii)$ The force on the floor by the system (action); the force on the system by the floor (reaction). In addition,for $(b)$,the force on the block by the cylinder and the force on the cylinder by the block also constitute an action-reaction pair.

(N/A) Zero net force: The rain drop is falling with a constant speed,meaning its acceleration is zero. According to Newton's second law of motion,the net force acting on the rain drop is zero.
$(b)$ Zero net force: The weight of the cork acts downward and is balanced by the buoyant force exerted by the water in the upward direction. Thus,the net force is zero.
$(c)$ Zero net force: The kite is stationary in the sky,meaning it has no acceleration. According to Newton's first law of motion,the net force acting on the kite is zero.
$(d)$ Zero net force: The car is moving with a constant velocity,so its acceleration is zero. According to Newton's second law of motion,the net force acting on the car is zero.
$(e)$ Zero net force: The high-speed electron is far from all material objects and external fields,so no force acts upon it.

(A) The net force is $0.5 \, N$ in the vertically downward direction in all cases.
Acceleration due to gravity $(g)$,irrespective of the direction of motion of an object,always acts downward. The gravitational force is the only force acting on the pebble in all three cases. Its magnitude is given by Newton's second law of motion as:
$F = m \times a$
Where,$F$ is the net force,$m$ is the mass of the pebble $(0.05 \, kg)$,and $a = g = 10 \, m/s^2$.
Therefore,$F = 0.05 \times 10 = 0.5 \, N$.
The net force on the pebble in all three cases is $0.5 \, N$ and this force acts in the downward direction.
If the pebble is thrown at an angle of $45^{\circ}$ with the horizontal,it will have both horizontal and vertical components of velocity. At the highest point,only the vertical component of velocity becomes zero. However,the pebble will have a horizontal component of velocity throughout its motion. This component of velocity has no effect on the net force acting on the pebble,which remains $0.5 \, N$ downwards.

(A) $1\; N$; vertically downward. Mass of the stone,$m = 0.1\; kg$. Acceleration of the stone,$a = g = 10\; m/s^2$. As per Newton's second law of motion,the net force acting on the stone is $F = ma = mg = 0.1 \times 10 = 1\; N$. Acceleration due to gravity always acts in the downward direction.
$(b)$ $1\; N$; vertically downward. The train is moving with a constant velocity,so its acceleration is zero. Once the stone is dropped,no horizontal force acts on it. The only force acting is gravity,which is $1\; N$ vertically downward.
$(c)$ $1\; N$; vertically downward. Once the stone is dropped,it is no longer in contact with the train. The horizontal force due to the train's acceleration stops acting on the stone. The only force remaining is gravity,which is $1\; N$ vertically downward.
$(d)$ $0.1\; N$; in the direction of motion of the train. The stone is at rest relative to the train,so it shares the train's acceleration $a = 1\; m/s^2$. The net force is $F = ma = 0.1 \times 1 = 0.1\; N$,acting in the direction of the train's motion.

(N/A) In order to pull a cart,a horse pushes the ground backward with some force. The ground in turn exerts an equal and opposite reaction force upon the feet of the horse. This reaction force causes the horse to move forward. An empty space is devoid of any such reaction force. Therefore,a horse cannot pull a cart and run in empty space.
$(b)$ When a speeding bus stops suddenly,the lower portion of a passenger's body,which is in contact with the seat,suddenly comes to rest. However,the upper portion tends to remain in motion (as per the first law of motion). As a result,the passenger's upper body is thrown forward in the direction in which the bus was moving.
$(c)$ While pulling a lawn mower,a force at an angle $\theta$ is applied on it. The vertical component of this applied force $(F \sin \theta)$ acts upward,which reduces the effective weight of the mower $(mg - F \sin \theta)$. On the other hand,while pushing a lawn mower,the vertical component of the applied force acts in the direction of the weight of the mower,increasing the effective weight $(mg + F \sin \theta)$. Since the effective weight is lesser in the first case,pulling the lawn mower is easier than pushing it.
$(d)$ According to Newton's second law of motion,$F = ma = m \frac{\Delta v}{\Delta t}$. The impact force $(F)$ is inversely proportional to the impact time $(\Delta t)$,i.e.,$F \propto \frac{1}{\Delta t}$. By moving his hands backward,a cricketer increases the time of impact $(\Delta t)$,which results in a decrease in the stopping force,thereby preventing the hands from getting hurt.

(N/A) Mass of the man,$m = 65 \; kg$.
Acceleration of the belt,$a = 1 \; m/s^2$.
Coefficient of static friction,$\mu = 0.2$.
The net force $F$ acting on the man is given by Newton's second law of motion as:
$F = m \times a = 65 \times 1 = 65 \; N$.
The man will continue to be stationary with respect to the conveyor belt until the net force required to accelerate the man is less than or equal to the maximum static frictional force $f_{s,max}$ exerted by the belt.
$f_{s,max} = \mu \times N = \mu \times m \times g$.
Let $a'$ be the maximum acceleration of the belt. For the man to remain stationary:
$m \times a' \leq \mu \times m \times g$
$a' \leq \mu \times g$
$a' \leq 0.2 \times 10 = 2 \; m/s^2$.
Therefore,the maximum acceleration of the belt up to which the man can stand stationary is $2 \; m/s^2$.

(A-D) Mass of the helicopter,$m_h = 1000 \; kg$. Mass of the crew and passengers,$m_p = 300 \; kg$.
Total mass of the system,$m = 1300 \; kg$.
Acceleration of the helicopter,$a = 15 \; m s^{-2}$.
$(a)$ Using Newton's second law of motion for the crew and passengers,the normal force $R$ exerted by the floor is:
$R - m_p g = m_p a \implies R = m_p(g + a)$
$R = 300(10 + 15) = 300 \times 25 = 7500 \; N$.
By Newton's third law,the force on the floor by the crew and passengers is $7500 \; N$,directed downward.
$(b)$ Using Newton's second law for the entire system (helicopter + crew + passengers):
$F_{air} - mg = ma \implies F_{air} = m(g + a)$
$F_{air} = 1300(10 + 15) = 1300 \times 25 = 32500 \; N$.
By Newton's third law,the action of the rotor on the surrounding air is $32500 \; N$,directed downward.
$(c)$ The force on the helicopter due to the surrounding air is $32500 \; N$,directed upward.

(N/A) Mass of body $A$,$m_{A} = 5 \, kg$.
Mass of body $B$,$m_{B} = 10 \, kg$.
Applied force,$F = 200 \, N$.
Coefficient of friction,$\mu = 0.15$.
The total force of friction is given by $f_{s} = \mu(m_{A} + m_{B})g = 0.15(5 + 10) \times 10 = 22.5 \, N$ (leftward).
$(a)$ The net force acting on the partition is $F_{net} = F - f_{s} = 200 - 22.5 = 177.5 \, N$ (rightward). By Newton's third law,the reaction of the wall is $177.5 \, N$ (leftward).
$(b)$ The force of friction on mass $A$ is $f_{A} = \mu m_{A} g = 0.15 \times 5 \times 10 = 7.5 \, N$ (leftward). The net force exerted by $A$ on $B$ is $F_{AB} = F - f_{A} = 200 - 7.5 = 192.5 \, N$ (rightward). By Newton's third law,$B$ exerts $192.5 \, N$ on $A$ (leftward).
When the wall is removed,the system moves with acceleration $a = \frac{F - f_{s}}{m_{A} + m_{B}} = \frac{177.5}{15} \approx 11.83 \, m/s^{2}$.
The force exerted by $A$ on $B$ becomes $F'_{AB} = m_{B} a + f_{B} = 10 \times 11.83 + (0.15 \times 10 \times 10) = 118.3 + 15 = 133.3 \, N$. Yes,the answer to $(b)$ changes when the bodies are in motion.

(A) Mass of the block,$m = 15 \; kg$.
Coefficient of static friction,$\mu = 0.18$.
Acceleration of the trolley,$a = 0.5 \; m s^{-2}$.
According to Newton's second law of motion,the force $F$ required to move the block with the trolley is $F = ma = 15 \times 0.5 = 7.5 \; N$.
The maximum force of static friction available is $f_{max} = \mu mg = 0.18 \times 15 \times 9.8 = 26.46 \; N$ (taking $g = 9.8 \; m s^{-2}$). Even if we take $g = 10 \; m s^{-2}$,$f_{max} = 27 \; N$.
Since the required force $(7.5 \; N)$ is less than the maximum static friction $(27 \; N)$,the block will not slide on the trolley.
$(a)$ For a stationary observer on the ground,the block accelerates along with the trolley at $0.5 \; m s^{-2}$ for the first $20 \; s$ and then moves with a uniform velocity of $v = at = 0.5 \times 20 = 10 \; m s^{-1}$.
$(b)$ For an observer moving with the trolley,the block remains at rest because the frictional force provides the necessary acceleration to keep it stationary relative to the trolley.

(C) Mass of the block,$m = 1\; kg$.
Spring constant,$k = 100\; N m^{-1}$.
Displacement of the block,$x = 10\; cm = 0.1\; m$.
According to the work-energy theorem,the work done by all forces is equal to the change in kinetic energy. Since the block starts from rest and comes to rest,the change in kinetic energy is zero.
Work done by gravity + Work done by spring + Work done by friction = $0$.
$mgx \sin 37^{\circ} - \frac{1}{2} k x^2 - f x = 0$.
Here,$f = \mu N = \mu mg \cos 37^{\circ}$.
Substituting the values:
$(1)(9.8)(0.1) \sin 37^{\circ} - \frac{1}{2} (100)(0.1)^2 - \mu (1)(9.8)(0.1) \cos 37^{\circ} = 0$.
$0.98 \times 0.6018 - 0.5 - \mu (0.98)(0.7986) = 0$.
$0.5898 - 0.5 = \mu (0.7826)$.
$0.0898 = \mu (0.7826)$.
$\mu = \frac{0.0898}{0.7826} \approx 0.115$.

(D) In daily life,motion is initiated or controlled by the application of an external force,which is a fundamental concept of $Newton's$ laws of motion.
$1$. $A$ stationary football is kicked: The force applied by the foot changes the state of rest of the ball.
$2$. To throw a stone in an upward direction,one has to apply an upward push (force).
$3$. Branches of a tree swing due to wind: The moving air exerts a force on the branches,causing them to move.
$4$. $A$ boat moves in a flowing river without anyone rowing it: The force exerted by the flowing water moves the boat.
All these examples demonstrate how external forces control or initiate motion.

(N/A) Friction is like fire; sometimes it is desirable and necessary,and sometimes it is undesirable.
Advantages of Friction:
$(1)$ It allows us to walk on the ground without slipping.
$(2)$ It enables us to hold objects firmly.
$(3)$ It allows us to write on surfaces like paper or boards.
$(4)$ It helps in transmitting motion between wheels via conveyor belts.
Disadvantages of Friction:
$(1)$ It opposes relative motion and dissipates mechanical energy in the form of heat.
$(2)$ It causes wear and tear in the moving parts of machines.
$(3)$ It leads to the wearing out of clothes,shoes,and vehicle tires.
Remedies to Reduce Friction:
$(1)$ Lubricants: Using oils or greases between moving parts reduces kinetic friction.
$(2)$ Ball-bearings: Using ball-bearings between moving parts converts sliding friction into rolling friction,which is significantly lower.
$(3)$ Air Cushion: Maintaining a thin cushion of air between solid surfaces can effectively reduce friction,as illustrated in the figure.

(N/A) The coefficients of friction are denoted by $\mu_s$ (static friction),$\mu_k$ (kinetic friction),and $\mu_r$ (rolling friction).
For a given pair of surfaces,the magnitude of these coefficients generally follows the inequality:
$\mu_s > \mu_k > \mu_r$.
This relationship exists because static friction must overcome the interlocking of surface irregularities at rest,kinetic friction involves sliding where the surfaces have less time to interlock,and rolling friction is significantly lower as it involves minimal deformation and contact area changes.

(N/A) We can solve different problems of dynamics using Newton's three laws of motion.
Sometimes,a problem involves more than one body,and these bodies exert forces on each other.
Moreover,every body experiences a gravitational force. To solve such problems,we must define a 'system' whose motion is to be discussed,while the remaining parts and external agencies are considered the 'environment'.
To solve the problem,follow these steps:
$(1)$ Draw a schematic diagram showing the assembly of different objects,their connections,and supports.
$(2)$ Select the object as the system whose motion you want to analyze. If the system consists of multiple objects,ensure that the acceleration vector of all these objects is the same.
$(3)$ List all forces acting on the system by the environment. Do not include internal forces acting within the system.
$(4)$ Represent the system as a point and draw all external forces acting on it as vectors originating from that point. This is called a Free Body Diagram $(FBD)$. This does not mean the system is free from forces,but only that the forces acting $ON$ the system are shown.
$(5)$ If necessary,select another system and repeat the steps. Use Newton's third law: for example,if the force exerted by $B$ on $A$ is $\vec{F}$ in the $FBD$ of $A$,then the force exerted by $A$ on $B$ in the $FBD$ of $B$ is $-\vec{F}$.

(A) The component of the force $F$ in the horizontal direction ($x$-direction) is given by $F_x = F \cos(60^{\circ})$.
Given $F = 36 \, \text{dyne}$ and $\theta = 60^{\circ}$.
$F_x = 36 \times \cos(60^{\circ}) = 36 \times 0.5 = 18 \, \text{dyne}$.
According to Newton's second law,the acceleration in the horizontal direction $a_x$ is given by $a_x = \frac{F_x}{m}$.
Given mass $m = 18 \, \text{g}$.
$a_x = \frac{18 \, \text{dyne}}{18 \, \text{g}} = 1 \, \text{cm/s}^2$.

(C) Work done by a force is given by $W = \vec{F} \cdot \vec{d} = Fd \cos(\theta)$,where $\theta$ is the angle between the force vector and the displacement vector.
In the case of kinetic friction,the frictional force always acts in the direction opposite to the relative motion (displacement) of the object. Thus,the angle $\theta$ between the frictional force and the displacement is $180^{\circ}$.
Since $\cos(180^{\circ}) = -1$,the work done by kinetic friction is always negative $(W = -Fd)$.
However,in the case of static friction,it acts in the direction of the intended motion or relative motion. For example,when a person walks,static friction acts in the direction of motion,making the work done by static friction positive. Therefore,the work done by frictional force can be positive or negative depending on the type of friction and the context.

(B) When a projectile explodes in mid-air,the explosion is caused by internal forces (such as chemical energy release).
According to Newton's laws,internal forces do not change the motion of the centre of mass of a system.
The only external force acting on the projectile throughout its flight,including during and after the explosion,is the gravitational force $(F_{ext} = mg)$.
Therefore,the centre of mass of the projectile continues to follow the original parabolic trajectory determined by the external gravitational force.

(D) The contact force $R$ between two surfaces is the resultant of the normal force $N$ and the frictional force $f$.
By definition,the normal force $N$ acts perpendicular to the surface,and the frictional force $f$ acts parallel to the surface.
Since these two forces are mutually perpendicular,the magnitude of the resultant contact force $R$ is given by the vector sum:
$R = \sqrt{N^2 + f^2}$.
For an object on an inclined plane,$N = mg \cos \theta$ and $f = \mu N$ (if sliding) or $f = mg \sin \theta$ (if in equilibrium).
Thus,the general expression for the contact force is $R = \sqrt{N^2 + f^2}$.

(B) Mass of the gun,$m_1 = 100 \, kg$.
Mass of the ball,$m_2 = 1 \, kg$.
Height of the cliff,$h = 500 \, m$.
Horizontal distance travelled by the ball,$x = 400 \, m$.
Acceleration due to gravity,$g = 10 \, m/s^2$.
First,we calculate the time taken by the ball to reach the ground using the vertical motion equation:
$h = \frac{1}{2} g t^2$
$500 = \frac{1}{2} \times 10 \times t^2$
$500 = 5 t^2$
$t^2 = 100$
$t = 10 \, s$.
Next,we find the horizontal velocity $(u)$ of the ball:
$x = u t$
$400 = u \times 10$
$u = 40 \, m/s$.
Using the law of conservation of linear momentum,the momentum of the gun must be equal and opposite to the momentum of the ball:
$m_1 v = m_2 u$
$100 \times v = 1 \times 40$
$v = \frac{40}{100} = 0.4 \, m/s$.
Therefore,the recoil velocity of the gun is $0.4 \, m/s$.

(D) The frictional force provides the necessary centripetal force for circular motion.
For a circular path of radius $r$,the maximum speed $v$ is given by $\frac{mv^2}{r} = \mu mg$,which simplifies to $v = \sqrt{\mu rg}$.
$1$. For path $ABC$ (radius $2R = 200 \, m$):
Length $L_1 = \frac{3}{4} \times (2\pi \times 2R) = 3\pi R = 300\pi \, m$.
Speed $v_1 = \sqrt{0.1 \times 200 \times 10} = \sqrt{200} \approx 14.14 \, m/s$.
Time $t_1 = \frac{300\pi}{14.14} \approx 66.66 \, s$.
$2$. For path $DEF$ (radius $R = 100 \, m$):
Length $L_2 = \frac{1}{4} \times (2\pi R) = \frac{\pi R}{2} = 50\pi \, m$.
Speed $v_2 = \sqrt{0.1 \times 100 \times 10} = 10 \, m/s$.
Time $t_2 = \frac{50\pi}{10} = 5\pi \approx 15.71 \, s$.
$3$. For straight paths $CD$ and $FA$ (length $R = 100 \, m$ each):
Total length $L_3 = R + R = 200 \, m$.
Speed $v_3 = 50 \, m/s$.
Time $t_3 = \frac{200}{50} = 4.0 \, s$.
Total time $T = t_1 + t_2 + t_3 = 66.66 + 15.71 + 4.0 = 86.37 \, s$.

(A) Given:
Mass of helicopter $m_{1} = 2000 \,kg$
Mass of crew and passengers $m_{2} = 500 \,kg$
Acceleration $a = 15 \,m s^{-2}$
Acceleration due to gravity $g = 10 \,m s^{-2}$
$(a)$ The force on the floor of the helicopter by the crew and passengers is the apparent weight $F_{1} = m_{2}(g + a)$.
$F_{1} = 500 \times (10 + 15) = 500 \times 25 = 12500 \,N$ (downward).
$(b)$ The action of the rotor on the surrounding air is equal to the total force required to lift the entire system (helicopter + crew + passengers) with acceleration $a$.
$F_{2} = (m_{1} + m_{2})(g + a) = (2000 + 500) \times (10 + 15) = 2500 \times 25 = 62500 \,N$ (downward).
$(c)$ By Newton's third law,the force on the helicopter due to the surrounding air is the reaction to the action force $F_{2}$.
$F_{3} = 62500 \,N$ (upward).