Laws of Motion

$\begin{aligned} \mathrm{f}_{\mathrm{K}} & =\mu \mathrm{mg} \cos \theta \\ & =0.1 \times \frac{50 \times \sqrt{3}}{2} \\ & =2.5 \sqrt{3} \mathrm{~N} \end{aligned}$

$\begin{aligned} \mathrm{F}_1= & \mathrm{mg} \sin \theta+\mathrm{f}_{\mathrm{K}} \\ = & 25+2.5 \sqrt{3} \end{aligned}$

$\begin{aligned} \mathrm{F}_2= & \mathrm{mg} \sin \theta-\mathrm{f}_{\mathrm{K}} \\ & =25-2.5 \sqrt{3} \\ \therefore \mathrm{F}_1 & -\mathrm{F}_2=5 \sqrt{3} \mathrm{~N} \end{aligned}$

$\begin{aligned} & a=\frac{\left(m_1-m_2\right) g}{\left(m_1+m_2\right)}=\frac{g}{8} \\ & 8 m_1-8 m_2=m_1+m_2 \\ & 7 m_1=9 m_2 \\ & \frac{m_1}{m_2}=\frac{9}{7} \end{aligned}$

For M block

$\begin{aligned} & 10 \mathrm{~g} \sin 53^{\circ}-\mu(10 \mathrm{~g}) \cos 53^{\circ}-\mathrm{T}=10 \times 2 \\ & \mathrm{~T}=80-15-20 \\ & \mathrm{~T}=45 \mathrm{~N} \end{aligned}$

For $\mathrm{m}$ block

$\begin{aligned} & \mathrm{T}-\mathrm{mg} \sin 37^{\circ}-\mu \mathrm{mg} \cos 37^{\circ}=\mathrm{m} \times 2 \\ & 45=10 \mathrm{~m} \\ & \mathrm{~m}=4.5 \mathrm{~kg} \end{aligned}$

Given:

The equation of the surface: $y = \frac{x^2}{4}$.

Coefficient of friction: $\mu = 0.5$.

Gravitational acceleration: $g$ (assumed constant).

1. Slope of the Surface:

The slope of the surface at any point is given by:

$ \tan\theta = \frac{dy}{dx} $

From the equation of the surface:

$ y = \frac{x^2}{4} $

Differentiating with respect to $x$:

$ \frac{dy}{dx} = \frac{x}{2} $

Thus, the slope at any point is:

$ \tan\theta = \frac{x}{2} $

2. Forces Acting on the Block:

At the point where the block is placed:

Weight of the block acts vertically downward: $ mg $.

Normal force acts perpendicular to the surface.

Frictional force acts parallel to the surface, opposing the component of the weight that causes slipping.

3. Condition for No Slipping:

The block will not slip if the frictional force is sufficient to counteract the component of the gravitational force parallel to the slope. The frictional force is:

$ f = \mu N $

The normal force $N$ is given by:

$ N = mg \cos\theta $

The component of weight parallel to the slope is:

$ F_{\text{parallel}} = mg \sin\theta $

For the block to not slip:

$ f \geq F_{\text{parallel}} $

Substitute $f = \mu N$ and $N = mg \cos\theta$:

$ \mu (mg \cos\theta) \geq mg \sin\theta $

Simplify:

$ \mu \cos\theta \geq \sin\theta $

Divide through by $\cos\theta$:

$ \mu \geq \tan\theta $

4. Maximum Slope Without Slipping:

From the above condition:

$ \tan\theta \leq \mu $

Substitute $\mu = 0.5$:

$ \tan\theta \leq 0.5 $

Thus:

$ \frac{x}{2} \leq 0.5 $

$ x \leq 1 $

5. Maximum Height:

The height $y$ of the block at $x = 1$ is obtained from the surface equation:

$ y = \frac{x^2}{4} $

Substitute $x = 1$:

$ y = \frac{1^2}{4} = \frac{1}{4} \, \text{m} $

Final Answer:

The maximum height above the ground at which the block can be placed without slipping is:

$ \boxed{\frac{1}{4} \, \text{m}} $

Thus, the correct option is Option D.

$a_A=a_B=a_c=\frac{F}{5+3+2}=\frac{80}{10}=8 \mathrm{~m} / \mathrm{s}^2$

$\mathrm{T_1=5\times8=40}$

$\mathrm{T_2-T_1=3\times8\Rightarrow T_2=64}$

$\begin{aligned} & 40-2 \mathrm{~T}=4 \mathrm{a} \\\\ & \mathrm{T}-10=4 \mathrm{a} \Rightarrow 20=12 \mathrm{a} \\\\ & \Rightarrow \mathrm{a}=5 / 3 \Rightarrow 2 \mathrm{a}=\frac{g}{3} \end{aligned}$

Let's analyze both statements:

Statement (I): The limiting force of static friction depends on the area of contact and independent of materials.

This statement is incorrect. The limiting force of static friction does not depend on the area of contact but is dependent on the materials in contact. According to the law of static friction, the maximum static frictional force $ f_s $ that can occur before motion commences is given by the product of the coefficient of static friction $ \mu_s $ and the normal reaction force $ N $:

$ f_s = \mu_s \times N $

The coefficient $ \mu_s $ is a property that depends on the materials in contact, not on the area of contact. The normal force $ N $ is the force perpendicular to the surfaces in contact, influenced by the weight of the object and any other perpendicular forces acting upon it.

Statement (II): The limiting force of kinetic friction is independent of the area of contact and depends on materials.

This statement is correct. Once an object is in motion, the kinetic frictional force $ f_k $ opposing its motion is given by the product of the coefficient of kinetic friction $ \mu_k $ and the normal force $ N $:

$ f_k = \mu_k \times N $

The coefficient $ \mu_k $, like $ \mu_s $, is also a property dependent on the materials of the surfaces in contact. It generally has a lower value than $ \mu_s $, which is why objects tend to be easier to keep moving once they've started. The kinetic frictional force is independent of the area of contact between the two surfaces.

Given these explanations, the correct answer would be:

Option A: Statement I is incorrect but Statement II is correct.

Given:

Area of cross-section of the pipe, $ A = 5 \, \text{cm}^2 = 5 \times 10^{-4} \, \text{m}^2 $

Velocity of water, $ v = 5 \, \text{m/s} $

Density of water, $ \rho = 1 \, \text{g/cm}^3 = 1000 \, \text{kg/m}^3 $

To find the mass flow rate $\left(\frac{dm}{dt}\right)$, we use the formula:

$ \frac{dm}{dt} = \rho \cdot A \cdot v $

Calculating, we get:

$ \frac{dm}{dt} = 1000 \times 5 \times 10^{-4} \times 5 = 2.5 \, \text{kg/s} $

The water hits the block and rebounds with the same velocity but in the opposite direction. The change in velocity $(\Delta v)$ is:

$ \Delta v = v_{\text{final}} - v_{\text{initial}} = -5 - 5 = -10 \, \text{m/s} $

The force exerted by the water stream $F$ is calculated using the rate of change of momentum:

$ F = \frac{dm}{dt} \cdot \Delta v = 2.5 \times (-10) = -25 \, \text{N} $

(Note: The negative sign indicates the direction; for magnitude, we use the positive value.)

Thus, the force is $F = 25 \, \text{N}$.

Using Newton's Second Law:

$ F = ma \quad \Rightarrow \quad a = \frac{F}{m} = \frac{25}{5} = 5 \, \text{m/s}^2 $

Therefore, the initial acceleration of the block is $5 \, \text{m/s}^2$.

From net force acting on the body, From figure (i) and figure (ii)

$ \begin{array}{l} 20-f=m a \\\\ 20-f=m \times 6 \\\\ 30-f=m \times 11 \end{array} $

where $ f $ is frictional force (kinetic) From Eqs. (i) and (ii)

$ \begin{aligned} m & =2 \mathrm{~kg} \text { and } f=8 \mathrm{~N} \\\\ f & =\mu_{k} m g \\\\ \mu_{k} & =\frac{f}{m g}=\frac{8}{2 \times 10} \\\\ \mu_{k} & =0.4 \end{aligned} $

$ \mu_{k} $ coefficient of kinetic energy.

$ \text { Case-1 } M g-F_B=M a_0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Case-2

Let removed mass is $m$.

$ \begin{aligned} &\therefore \quad F_B-(M-m) g=(M-m) 2 a_0 \,\,\,\,\,\,\,\,\,\,\,\,\,\ldots(ii)\\ &\text { Adding Eqs. (i) and (ii), we get }\\ &\begin{aligned} & M g-(M-m) g=M a_0+(M-m) \cdot 2 a_0 \\ & \Rightarrow \quad g[M-M+m]=a_0(M+2 M-2 m) \\ & \Rightarrow \quad g m=a_0(3 M-2 m) \\ & \Rightarrow \quad g m+2 m a_0=3 M a_0 \\ & \Rightarrow \quad m\left(g+2 a_0\right)=3 M a_0 \Rightarrow m=\frac{3 M a_0}{g+2 a_0} \end{aligned} \end{aligned} $

According to figure, the net force $F_{\text {net }}$ acting downward in inclined plane is the difference between the component of the gravitational force parallel to the plane and the frictional force.

$ F_{\text {net }}=m g \sin \theta-\mu m g \cos \theta $

Using Newton's second law,

$ \begin{aligned} F_{\mathrm{net}} & =m \cdot a \\ a & =\frac{F_{\mathrm{net}}}{m} \end{aligned} $

Putting value of $F_{\text {net }}$,

$ \begin{aligned} & a=\frac{m g \sin \theta-\mu m g \cos \theta}{m} \\ & a=g \sin \theta-\mu g \cos \theta \\ & a=g(\sin \theta-\mu \cos \theta) \end{aligned} $

To calculate the external force required to push a 2 kg mass up an inclined plane at an acceleration of $4 \, \text{ms}^{-2}$, we first need to understand the forces at play when the body moves down the plane.

When sliding down:

The net force ($F_{\text{net}}$) is given by:

$ F_{\text{net}} = mg \sin \theta - f $

Applying Newton's second law $F_{\text{net}} = ma$, we have:

$ ma = mg \sin \theta - f $

Given values:

$m = 2 \, \text{kg}$

$a = 4 \, \text{ms}^{-2}$

$g = 10 \, \text{ms}^{-2}$

$\theta = 30^\circ$

By substituting these values:

$ 2 \times 4 = 2 \times 10 \times \sin 30^\circ - f $

Evaluating further gives:

$ 8 = 20 \times 0.5 - f $

$ f = 10 - 8 = 2 \, \text{N} $

When the body is sliding up the inclined plane, the required force $F$ is determined by:

Modifying the net force equation for the upward motion, we have:

$ F - mg \sin \theta - f = ma $

Rearranging gives:

$ F = ma + mg \sin \theta + f $

Substituting the values:

$ F = 2 \times 4 + 2 \times 10 \times \sin 30^\circ + 2 $

$ F = 8 + 10 + 2 = 20 \, \text{N} $

Therefore, the force required to take the body up with the same acceleration is $20 \, \text{N}$.

Given:

Mass of the gun, $ M = 100 \, \text{kg} $

Mass of the ball, $ m = 1 \, \text{kg} $

Height of the cliff, $ h = 500 \, \text{m} $

Distance from the cliff to where the ball lands, $ s = 400 \, \text{m} $

Acceleration due to gravity, $ g = 10 \, \text{m/s}^2 $

Step 1: Calculate Time of Flight

The time taken by the ball to fall to the ground can be determined using the formula for the time of free fall:

$ t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 500}{10}} = \sqrt{100} = 10 \, \text{s} $

Step 2: Determine Horizontal Velocity

The horizontal distance covered by the ball is given by $ s = u \times t $, where $ u $ is the horizontal velocity of the ball:

$ \begin{aligned} s &= u \times t \\ 400 &= u \times 10 \\ u &= 40 \, \text{m/s} \end{aligned} $

Step 3: Apply Conservation of Linear Momentum

Initially, the system (cannon + ball) is at rest, so the initial momentum is zero. After the ball is fired, by the conservation of linear momentum, the total momentum should still be zero:

$ 0 = Mv + mu $

Solving for the recoil velocity $ v $ of the gun, we have:

$ \begin{aligned} Mv &= -mu \\ v &= -\frac{mu}{M} = -\frac{(1) \times (40)}{100} \\ v &= -0.4 \, \text{m/s} \end{aligned} $

The negative sign indicates that the recoil velocity of the gun is opposite in direction to the velocity of the ball. Thus, the recoil velocity of the gun is $ 0.4 \, \text{m/s} $.

To find the acceleration of the block, we start with the given values:

Mass of the block, $ m = 5 \, \text{kg} $

Coefficient of friction, $ \mu = 0.5 $

Horizontal force applied, $ F_H = 60 \, \text{N} $

Acceleration due to gravity, $ g = 10 \, \text{m/s}^2 $

First, calculate the frictional force ($ F_f $) using the formula:

$ F_f = \mu \times N = \mu \times m \times g $

Substitute the values:

$ F_f = 0.5 \times 5 \times 10 $

$ F_f = 25 \, \text{N} $

Now, compute the net force ($ F_{\text{net}} $) acting on the block by subtracting the frictional force from the horizontal force:

$ F_{\text{net}} = F_H - F_f $

$ F_{\text{net}} = 60 - 25 = 35 \, \text{N} $

Finally, calculate the acceleration ($ a $) of the block using Newton's second law:

$ a = \frac{F_{\text{net}}}{m} $

$ a = \frac{35}{5} $

$ a = 7 \, \text{m/s}^2 $

Therefore, the acceleration of the block is $ 7 \, \text{m/s}^2 $.

Given the following information:

Mass of the block ($m$) = 4 kg

Rate of water release $\left(\frac{dm}{dt}\right)$ = 2 kg/s

Speed of the water ($v$) = 10 m/s

To find the acceleration of the block, start by calculating the force exerted on it. The force ($F$) can be calculated using the relationship:

$ F = \frac{dm}{dt} \cdot v $

Substitute the given values:

$ F = 2 \times 10 = 20 \, \text{N} $

Next, use Newton's second law to find the acceleration ($a$) of the block. The relationship between force, mass, and acceleration is:

$ F = m \cdot a $

Rearranging for acceleration gives:

$ a = \frac{F}{m} $

Plug in the calculated force and the mass of the block:

$ a = \frac{20}{4} = 5 \, \text{m/s}^2 $

Therefore, the acceleration of the block is $5 \, \text{m/s}^2$.

Given:

The angle of inclination, $ \theta = 30^\circ $.

The speed of the conveyor belt, $ v_{\text{belt}} = \sqrt{\frac{g h}{6}} $.

The coefficient of friction, $ \mu = \frac{5}{3\sqrt{3}} $.

Calculation of Maximum Possible Acceleration:

The maximum force of friction that can act on the person while climbing is given by:

$ F_{\max} = \mu m g \cos \theta $

The component of gravitational force acting parallel to the incline is:

$ F_{\text{parallel}} = m g \sin \theta $

Using Newton’s second law, the maximum acceleration $ a $ is:

$ a = \frac{F_{\max} - F_{\text{parallel}}}{m} $

Substitute the given values:

$ \begin{aligned} a & = \frac{\mu m g \cos \theta - m g \sin \theta}{m} \\ & = \frac{\frac{5}{3\sqrt{3}} m g \times \frac{\sqrt{3}}{2} - m g \times \frac{1}{2}}{m} \\ & = \frac{\frac{5}{6} m g - \frac{1}{2} m g}{m} \\ & = \frac{g}{3} \end{aligned} $

Solving for Time $ t $:

The displacement from point $ A $ to point $ B $ is $ AB = \frac{OB}{\sin \theta} = \frac{h}{1/2} = 2h $. The equation of motion is:

$ AB = v_{\text{belt}} \cdot t + \frac{1}{2} a t^2 $

Substituting the known values, we have:

$ \begin{aligned} 2h & = \sqrt{\frac{g h}{6}} \cdot t + \frac{1}{2} \cdot \frac{g}{3} \cdot t^2 \\ & = \sqrt{\frac{g h}{6}} \cdot t + \frac{g}{6} \cdot t^2 \end{aligned} $

Reorganizing terms gives:

$ 2h = \sqrt{\frac{g h}{6}} \cdot t + \frac{g}{6} \cdot t^2 $

Solving this quadratic equation in terms of $ t $, we find:

$ t = \sqrt{\frac{6h}{g}} $

Thus, the time taken for the person to travel from $ A $ to $ B $ with maximum possible acceleration is:

$ t = \sqrt{\frac{6h}{g}} $

Given:

Flow of water: $ q = 1 \, \text{kg/s} $

Speed of water: $ v = 5 \, \text{m/s} $

Mass of block: $ m = 2 \, \text{kg} $

The rate of momentum transfer can be calculated as:

$ \frac{dp}{dt} = q \times v $

Substituting the given values:

$ \frac{dp}{dt} = 1 \times 5 = 5 \, \text{kg m/s}^2 $

This rate of change of momentum is equivalent to the force $ F $ exerted by the water jet on the block. Thus, we have:

$ F = \frac{dp}{dt} = 5 \, \text{N} $

From Newton's second law, the force is also expressed as:

$ F = m \times a $

Substituting the known values:

$ 2 \times a = 5 $

Solving for $ a $ (acceleration of the block):

$ a = \frac{5}{2} = 2.5 \, \text{m/s}^2 $

Therefore, the acceleration of the block is $ 2.5 \, \text{m/s}^2 $.

Drawing the FBD according to question.

Insect is at highest point, (let it be in equilibrium)

$ N=m g \cos \theta \quad \ldots \text { (i) } $

and $f=m g \sin \theta \quad \ldots \text { (ii) } $

For limiting case of static friction,

$ \begin{array}{cc} & f=\mu N \\ \Rightarrow & m g \sin \theta=\mu m g \cos \theta \quad \ldots \text { (iii) } \\ \therefore & \mu=\tan \theta \end{array} $

From above relation, $\cos \theta=\frac{1}{\sqrt{\mu^2+1}}$

$ \left[\text { Identity used, } \cos \theta=\frac{1}{\sqrt{\tan ^2 \theta+1}}\right] $

In $\triangle O X P$

$ \begin{array}{ll} & \cos \theta=\frac{R-h}{R} \\ \therefore & \frac{R-h}{R}=\frac{1}{\sqrt{u^2+1}} \\ \therefore & h=R\left[1-\frac{1}{\sqrt{\mu^2+1}}\right] \end{array} $

Given that the mass ratio of the two objects is $\frac{m_1}{m_2} = \frac{1}{4}$, when both objects are subjected to the same force, their acceleration relationship is given by:

$ m_1 a_1 = m_2 a_2 \implies \frac{m_2}{m_1} = \frac{a_1}{a_2} = \frac{4}{1} $

The relation between velocity, acceleration, and time is:

$ v = u + at $

For the objects starting from rest:

$ v_1 = 0 + a_1 t_1 = a_1 t_1 $

$ v_2 = 0 + a_2 t_2 = a_2 t_2 $

Given that the kinetic energies are equal:

$ K_1 = K_2 \implies \frac{1}{2} m_1 v_1^2 = \frac{1}{2} m_2 v_2^2 $

Substituting the velocities:

$ \frac{1}{2} m_1 (a_1 t_1)^2 = \frac{1}{2} m_2 (a_2 t_2)^2 $

This simplifies to:

$ \frac{m_1}{m_2} \left(\frac{a_1}{a_2}\right)^2 = \left(\frac{t_2}{t_1}\right)^2 $

Substituting the given ratios:

$ \frac{1}{4} \times \left(\frac{4}{1}\right)^2 = \left(\frac{t_2}{t_1}\right)^2 $

This results in:

$ \frac{t_2}{t_1} = 2 $

Given, Linear density of rope

$ =0.5 \mathrm{~kg} \mathrm{~m}^{-1} $

So, mass of rope $\Rightarrow m=\lambda \cdot l$

$ m=0.5 \times 3=1.5 \mathrm{~kg} $

$\therefore$ Total mass of system,

$ m=18.5+1.5=20 \mathrm{~kg} $

Acceleration, $a=\frac{F}{m}=\frac{40}{20}=2 \mathrm{~m} / \mathrm{s}^2$

Distance, $s=u t+\frac{1}{2} a t^2$

$ \begin{aligned} & s=\frac{1}{2} a t^2 \quad(\because u=0 \text { given }) \\ & \frac{2 \times 9}{2}=t^2 \Rightarrow t=3 \mathrm{~s} \end{aligned} $

Given:

Mass of the block ($ m $) = 1.5 kg

Initial velocity ($ u $) = 10 m/s

Distance traveled before coming to rest ($ s $) = 12.5 m

Acceleration due to gravity ($ g $) = 10 m/s$^2$

To find the coefficient of kinetic friction ($ \mu_k $), we first use the kinematic equation to determine the acceleration ($ a $):

$ v^2 = u^2 + 2as $

Since the block comes to rest, the final velocity ($ v $) is 0. Plugging in the values:

$ 0^2 = (10)^2 + 2 \times a \times 12.5 $

Solving for $ a $:

$ -100 = 25a \quad \Rightarrow \quad a = -4 \, \text{m/s}^2 $

The negative acceleration indicates that the block is decelerating due to friction. The force applied is equal to the frictional force, and we relate it to the coefficient of kinetic friction:

$ F = ma = -f $

Where $ f $ is the frictional force, which is the product of the coefficient of kinetic friction ($ \mu_k $) and the normal force ($ N $). Since the block is on a horizontal surface, the normal force ($ N $) equals $ mg $:

$ -f = \mu_k N \quad \Rightarrow \quad -\mu_k mg = ma $

Rearranging the equation to solve for $ \mu_k $:

$ a = -\mu_k g \quad \Rightarrow \quad \mu_k = \frac{a}{g} = \frac{4}{10} = 0.4 $

Therefore, the coefficient of kinetic friction between the surface and the block is $ 0.4 $.

According to figure, component of forces

$T-2 m g \sin 37^{\circ}=2 m a \quad \ldots \text { (i) } $

$10 m g \sin 53^{\circ}-T=10 \mathrm{ma} \quad \ldots \text { (ii) } $

Adding Eqs. (i) and (ii), we get

$10 \mathrm{mg} \sin 53^{\circ}-2 m g \sin 37^{\circ}=12 \mathrm{ma} $

$10 \times g \times \frac{4}{5}-2 \times g \times \frac{3}{5}=12 \cdot a$

$ \begin{aligned} \frac{34 g}{5} & =12 a \\ a & =\frac{34 g}{60} \\ a & =\frac{17}{30} g \end{aligned} $

Given,

$ \begin{aligned} & s=3 \mathrm{~m}, t=3 \mathrm{~s}, u=0(\text { set free }) \\ & s=u t+\frac{1}{2} a t^2 \\ & 3=0+\frac{1}{2} a \times 3^2 \\ & a=\frac{2}{3} \mathrm{~m} / \mathrm{s}^2 \end{aligned} $

Applying the result of Newton 2nd law,

$ F=m a $

So, $ \begin{gathered} m_1 g-T=m_1 a \quad \ldots \text { (i) } \\ T-m_2 g=m_2 a \quad \ldots \text { (ii) } \end{gathered} $

Adding Eq. (i) and Eq. (ii),

$ \begin{aligned} g\left(m_1-m_2\right) & =a\left(m_1+m_2\right) \\ 10\left(m_1-m_2\right) & =\frac{2}{3}\left(m_1+m_2\right) \end{aligned} $

$\begin{aligned} 15 m_1-15 m_2 & =m_1+m_2 \\ 14 m_1 & =16 m_2 \\ 7 m_1 & =8 m_2 \\ \frac{m_1}{m_2} & =\frac{8}{7}\end{aligned}$

Since the particle is initially at rest, the net force acting on it is zero. This means that the forces $F_1$, $F_2$, and $F_3$ are balanced. Given that $F_2$ and $F_3$ are acting perpendicularly, we can represent the balance of forces as follows:

$F_1 = \sqrt{F_2^2 + F_3^2}$

We can now calculate the equivalent force resulting from $F_2$ and $F_3$:

$F_1 = \sqrt{(8 \mathrm{~N})^2 + (6 \mathrm{~N})^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \mathrm{~N}$

Since the particle is initially at rest, when the force $F_1$ is removed, only $F_2$ and $F_3$ remain. The net force acting on the particle is the equivalent force resulting from $F_2$ and $F_3$, which we have calculated to be $10 \mathrm{~N}$.

Now, we can use Newton's second law to find the acceleration of the particle:

$F = ma$

Where:

• $F$ is the net force acting on the particle
• $m$ is the mass of the particle
• $a$ is the acceleration of the particle

Rearranging the equation to solve for $a$:

$a = \frac{F}{m}$

Plugging in the values for $F$ and $m$:

$a = \frac{10 \mathrm{~N}}{5 \mathrm{~kg}} = 2 \mathrm{~ms}^{-2}$

The acceleration of the particle when the force $F_1$ is removed is $2 \mathrm{~ms}^{-2}$, which corresponds to Option C.

Given the velocity vector of a particle $v = (2t \hat{i}+3 t^{2} \hat{j}) \, \text{ms}^{-1}$, the acceleration $a$ is the derivative of the velocity vector with respect to time. So, we have:

$a = \frac{dv}{dt} = (2 \hat{i} + 6t \hat{j}) \, \text{ms}^{-2}$.

At $t=1 \, \text{s}$, the acceleration $a$ is $(2 \hat{i} + 6 \hat{j}) \, \text{ms}^{-2}$.

According to Newton's second law, the force $F$ is equal to the mass $m$ times acceleration $a$. The mass $m$ is given as $500 \, \text{g}$, or equivalently, $0.5 \, \text{kg}$.

Therefore, the force $F$ on the particle at $t=1 \, \text{s}$ is:

$F = m \cdot a = 0.5 \cdot (2 \hat{i} + 6 \hat{j}) = (1 \hat{i} + 3 \hat{j}) \, \text{N}$.

So, the force acting on the particle at $t=1 \, \text{s}$ is $(\hat{i} + x \hat{j}) \, \text{N}$, where $x=3$.

Therefore, the answer is $x=3$.

As we know that impulse is given by

$I = \Delta P = F \times \Delta t$

or $ I=$ Area of $f-t$ graph

For fig (a)

$\rightarrow \mathrm{I}=\frac{1}{2} \times$ base $\times$ height

$ =\frac{1}{2} \times 0.5 \times 1=0.25 \mathrm{~N}-\mathrm{sec} . $

For fig (b),

$I=$ length $\times$ width

$ =2 \times 0.5=1 \mathrm{~N} \text {-sec } $

For fig (c),

I $=\frac{1}{2} \times$ base $\times$ height

$ =\frac{1}{2} \times 1 \times 0.75=0.375 \mathrm{~N} \text {-sec. } $

For fig (d),

$I=\frac{1}{2} \times$ base $\times$ height

$ =\frac{1}{2} \times 2 \times 0.5=0.5 \mathrm{~N}-\mathrm{sec} \text {. } $

Impulse is highest for that figure, whose area under $F$-t is maximum and i.e. figure(b)

Option (D) is correct.

Drawing the FBD of the point where $F$ is applied

$T' = \sqrt 3 g$

$ \Rightarrow T\cos 30^\circ = \sqrt 3 g$

$ \Rightarrow T = 2g = 20$ N

$m = 20$ kg

$t = 20$ sec.

Acceleration $ = {F \over {20}}$ m/s$^2$

$\therefore$ $v = u + at$

$v = 0 + \left( {{F \over {20}}} \right)(20)$

$ = F$ ms$^{-1}$

Now for next 10 sec.

$S=ut$

$50=F(10)$

$F=5$

Smooth case:

$a = g\sin 45^\circ = {9 \over {\sqrt 2 }}$

${t_1} = \sqrt {{{2L} \over a}} = \sqrt {{{2L} \over {g/\sqrt 2 }}} = \sqrt {{{2\sqrt 2 L} \over g}} $ ..... (1)

Rough case:

$a = g\sin 45^\circ - \mu g\cos 45^\circ $

$ = {g \over {\sqrt 2 }}(1 - \mu )$

${t_2} = \sqrt {{{2L} \over a}} = \sqrt {{{2\sqrt 2 L} \over {g(1 - \mu )}}} $ ..... (2)

From (1) to (2) and ${t_1} = {{{t_2}} \over n}$ we have

$\sqrt {{{2\sqrt 2 L} \over g}} = {1 \over n}\sqrt {{{2\sqrt 2 L} \over {g(1 - \mu )}}} \Rightarrow \mu = 1 - {1 \over {{n^2}}}$

Let's consider $T$ tension in string and a acceleration of blocks.

FBD of $4 \mathrm{~kg}$

$4 g \sin \theta-\mathrm{T}=4 \mathrm{a}$

$ \frac{4 g \sqrt{3}}{2}-T=4 a $ .........(1)

FBD of $1 \mathrm{~kg}$

$T-g \sin 30^{\circ}=a$

$T-\frac{g}{2}=a \quad\quad...(2)$

From (1) and (2), we get

${{4g\sqrt 3 } \over 2} - {g \over 2} = 5a$

$ \Rightarrow $ ${{g\left( {4\sqrt 3 - 1} \right)} \over 2} = 5a$

$ \Rightarrow $ $5a = {{10\left( {4\sqrt 3 - 1} \right)} \over 2}$

$ \Rightarrow $ $a = \left( {4\sqrt 3 - 1} \right)$ m/s²

By putting value of "a" in equation (2), we get

$T - {{10} \over 2} = \left( {4\sqrt 3 - 1} \right)$

$ \Rightarrow $ $T = \left( {4\sqrt 3 - 1} \right) + 5$ = $\left( {4\sqrt 3 + 4} \right)$

$ \Rightarrow $ $T=4(\sqrt{3} + 1) N$

Statement I says that when the weight of an elevator is balanced with the tension of its cable, it can move up or down with a uniform speed. This is true because the weight of the elevator is balanced by the tension in the cable, which allows it to move smoothly and at a constant speed.

Statement II says that the force exerted by the floor of an elevator on a person's foot is greater than their weight when the elevator goes down with increasing speed. This is false because the force exerted by the floor on a person's foot is equal to their weight, regardless of the speed of the elevator. The person's weight is a constant force and does not change with the speed of the elevator. The apparent weight of a person may change with the speed of the elevator, but this is due to the effects of acceleration and not an increase in the force exerted by the floor.

Given :

$ \vec{v} = \alpha(y \hat{x} + 2x \hat{y}) $

For the x-component, $ v_x = \alpha y $ :

$ \frac{dv_x}{dt} = \alpha \frac{dy}{dt} = \alpha v_y $

Given $ v_y = 2\alpha x $ :

$ \frac{dv_x}{dt} = \alpha (2\alpha x) = 2\alpha^2 x $

For the y-component, $ v_y = 2\alpha x $ :

$ \frac{dv_y}{dt} = 2\alpha \frac{dx}{dt} = 2\alpha v_x $

Given $ v_x = \alpha y $ :

$ \frac{dv_y}{dt} = 2\alpha (\alpha y) = 2\alpha^2 y $

Thus, combining the components :

$ \vec{a} = 2\alpha^2 x \hat{x} + 2\alpha^2 y \hat{y} $

Now, the force acting on the particle is given by :

$ \vec{F} = m\vec{a} $

$ \vec{F} = m\alpha(2 \alpha x \hat{x} + 2 \alpha y \hat{y}) $

$ \vec{F} = m\alpha^2(2x \hat{x} + 2y \hat{y}) $

This is equivalent to :

$ \vec{F} = 2m\alpha^2(x \hat{x} + y \hat{y}) $

The correct answer is Option A :

$ \vec{F} = 2m\alpha^2(x \hat{x} + y \hat{y}) $

Given,

Mass of man $=50 \mathrm{~kg}$

Acceleration due to gravity

$ g=10 \mathrm{~m} / \mathrm{s}^2 $

Acceleration of lift $($ downward $)=g$

The net acceleration in case of lift, when lift is going downward

$ a_{\mathrm{net}}=g-a $

where $a$ is acceleration of lift.

$ \begin{aligned} & a_{\mathrm{net}}=g-g \quad(a=g \text { given }) \\ & a_{\mathrm{net}}=0 \end{aligned} $

So, the apparent weight is

$ \begin{aligned} & w=m a_{\text {net }} \\ & w=m \times 0 \Rightarrow w=0 \end{aligned} $

Given,

Weight of man when lift is not moving

$ =75 \mathrm{~kg} $

Lift is going downward with acceleration

$ =g=10 \mathrm{~m} / \mathrm{s}^2 $

The net acceleration for lift system going down

$ a_{\mathrm{net}}=g-a $

where $a$ is acceleration of lift.

$ \begin{gathered} a_{\mathrm{net}}=g-g \quad[a=g] \\ a_{\mathrm{net}}=0 \end{gathered} $

So, the weight inside it, when moving downward

$ \begin{aligned} & w=m a_{\mathrm{net}} \\ & w=0 \end{aligned} $

Thus, weight is zero.

Considering $w_2>w_1$, FBD is given below

The system is moved with acceleration $g$ upwards.

For motion of mass $m_2$, we have

$ w_2+m_2 g-T=m_2 a $

$ \begin{aligned} \Rightarrow & w_2+w_2-T =m_2 a \\ \Rightarrow & 2 w_2-T =m_2 a \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

For motion of mass $m_1$, we have

$ \begin{array}{ll} & T-m_1 g-w_1=m_1 a \\ \Rightarrow & T-w_1-w_1=m_1 a \\ \Rightarrow & T-2 w_1=m_1 a \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

From Eqs. (i), (ii), we get

$ \begin{aligned} & 2 w_2-T+T-2 w_1=m_2 a+m_1 a \\ & \Rightarrow \quad 2 w_2-2 w_1=a\left(m_2+m_1\right) \end{aligned} $

Since, $w_1=m_1 g$ and $w_2=m_2 g$

$ \begin{aligned} & \Rightarrow \quad 2 w_2-2 w_1=\left(\frac{w_2}{g}+\frac{w_1}{g}\right) \times a \\ & \Rightarrow \quad a=g\left(\frac{2 w_2-2 w_1}{w_1+w_2}\right) \end{aligned} $

Putting the value of a in Eq. (ii), we get

$ \begin{array}{rlrl} T-2 w_1 & =m_1 a \\ \Rightarrow & T-2 w_1 =m_1 g\left(\frac{2 w_2-2 w_1}{w_1+w_2}\right) \\ \Rightarrow & T =m_1 g\left(\frac{2 w_2-2 w_1}{w_1+w_2}\right)+2 w_1 \\ \Rightarrow & T =\left(\frac{4 w_1 w_2}{w_1+w_2}\right) g \quad\left(\because w_1=M_1 g\right) \end{array} $

Given, weight of body, $w=50 \mathrm{~N}$

Force $=28.28 \mathrm{~N}$

frictional force, $f=28.25 \times \cos 45^{\circ}$

$ =28.25 \times \frac{1}{\sqrt{2}}=20 \mathrm{~N} $

Normal reaction,

$ \begin{array}{r} R=50-28.2 \sin 45^{\circ} \\ =50-20=30 \mathrm{~N} \end{array} $

Hence, option (b) is correct.

Given,

Initially the person is at rest, after that start walking toward east.

Person does not skid or slip.

We know that, when we walk, our feet are pushing backward and therefore friction opposes this motion of pushing.

So, the direction of friction is opposite to walking i.e., east direction.

As the person does not skid or slip or roll the friction force which is acting on him is static in nature.

Hence, friction force is static and toward east in nature.

Mass of the Body: $6 \, \text{kg}$

Initial Velocity ($u$): $4 \, \text{m/s}$

Final Velocity ($v$): $6 \, \text{m/s}$

Force Applied ($F$): $12 \, \text{N}$

We start by determining the acceleration ($a$) using Newton's second law, defined as:

$ F = m \cdot a $

Substituting the known values:

$ a = \frac{F}{m} = \frac{12}{6} = 2 \, \text{m/s}^2 $

Next, we use the first equation of motion to find the time ($t$) taken for the velocity change:

$ v = u + a \cdot t $

Rearranging for $t$:

$ t = \frac{v - u}{a} = \frac{6 - 4}{2} = 1 \, \text{s} $

To find the displacement ($s$), we use the second equation of motion:

$ s = u \cdot t + \frac{1}{2} a \cdot t^2 $

Substituting the known values:

$ s = 4 \times 1 + \frac{1}{2} \times 2 \times (1)^2 $

$ s = 4 + 1 = 5 \, \text{m} $

Thus, the displacement of the body is $5 \, \text{m}$.

Consider the FBD as given below.

Equations of motion,

$\begin{aligned} & a_A=\frac{F}{m}(\text { in }-x \text { direction }) \\ & a_B=\frac{F}{M}(\text { in }+x \text { direction }) \end{aligned}$

Thus, we have, relative accleration of $A$ w.r.t. $B$ as

$\begin{aligned} a_{A B} & =a_A-a_B=\frac{-F}{m}-\frac{F}{M} \\ & =-F\left(\frac{M+m}{M m}\right) \quad \text{(along} ~\mathrm{-x}-\text{direction)} \end{aligned}$

Initial relative velocities of $A$ w.r.t $B, u_{A B}=3 v_0$ using equation $v^2-u^2+2 a s$,

$\begin{aligned} 0 & =9 v_0^2-\frac{2 F(M+m)}{M m} s \\ \Rightarrow \quad s & =\frac{9 M m v_0^2}{2 F(M+m)} \end{aligned}$

Distance moved relative to $B, S_{A B}=\frac{9 M m v_0^2}{2 F(M+m)}$