Laws of Motion

The length of the incline (hypotenuse) is given by :

$ \cos \theta=\frac{L}{s} \Rightarrow s=\frac{L}{\cos \theta} $

Since the wedge is accelerating to the left with $\mathrm{a}_0$, we must work in the non-inertial frame of the wedge. This causes a pseudo-force ( $\mathrm{ma}_0$ ) acting on the block to the right.

We resolve all forces along the surface of the incline :

The net acceleration relative to the incline $\left(\mathrm{a}_{\mathrm{rel}}\right)$ is :

$ \mathrm{a}_{\mathrm{rel}}=\frac{\mathrm{mg} \sin \theta-\mathrm{ma}_0 \cos \theta}{\mathrm{~m}}=\mathrm{g} \sin \theta-\mathrm{a}_0 \cos \theta $

Using the kinematic equation for distance s starting from rest $(\mathrm{u}=0)$ :

$ \mathrm{s}=\frac{1}{2} \mathrm{a}_{\mathrm{rel}} \mathrm{t}^2 . $

Putting $\mathrm{s}=\mathrm{L} \cos \theta$ and $\mathrm{a}_{\text {rel }}=\mathrm{g} \sin \theta-\mathrm{a}_0 \cos \theta:$

$ \frac{L}{\cos \theta}=\frac{1}{2}\left(g \sin \theta-a_0 \cos \theta\right) t^2 $

$ \mathrm{t}^2=\frac{2 \mathrm{~L}}{\cos \theta\left(\mathrm{~g} \sin \theta-\mathrm{a}_0 \cos \theta\right)}=\frac{2 \mathrm{~L}}{\mathrm{~g} \sin \theta \cos \theta-\mathrm{a}_0 \cos ^2 \theta} $

Using the identity $2 \sin \theta \cos \theta=\sin 2 \theta$ and $\cos ^2 \theta=\frac{1+\cos 2 \theta}{2}$ :

$ t^2=\frac{4 L}{2 g \sin \theta \cos \theta-2 a_0 \cos ^2 \theta}=\frac{4 L}{g \sin 2 \theta-2 a_0\left(\frac{1+\cos \theta}{2}\right)} $

$\Rightarrow $ $t^2=\frac{4 L}{g \sin 2 \theta-a_0(1+\cos \theta)}$

$\Rightarrow \mathrm{t}=\sqrt{\frac{4 \mathrm{~L}}{\mathrm{~g} \sin 2 \theta-\mathrm{a}_0(1+\cos \theta)}}$

Hence, the correct option is (B).

The forces acting on the block are,

Component of weight down the plane (along x -axis) $=\mathrm{mg} \sin \theta=5 \times 10 \times \sin 30^{\circ}=25 \mathrm{~N}$

Normal Force, $\mathrm{N}=\mathrm{mg} \cos \theta$

Since the block moves down, friction acts up the plane.

$ \mathrm{f}_{\mathrm{k}}=\mu_{\mathrm{k}} \mathrm{~N}=\mu(\mathrm{mg} \cos \theta) $

Maximum frictional force :

$ \mathrm{f}_{\mathrm{k}}=\frac{\sqrt{3}}{2} \times\left(5 \times 10 \times \cos \left(30^{\circ}\right)\right) $

$\Rightarrow $ $ \mathrm{f}_{\mathrm{k}}=\frac{\sqrt{3}}{2} \times\left(50 \times \frac{\sqrt{3}}{2}\right)=\frac{3}{4} \times 50=37.5 \mathrm{~N} $

Since the upward frictional force ( 37.5 N ) is greater than the downward component of weight ( 25 N ), the block will not move down on its own. To make it move down at a constant velocity, an additional downward force (P) must be applied.

The equilibrium equation along the incline is :

Net Force Downward $=$ Net Force Upward

$ \mathrm{P}+\mathrm{F}_{\text {weight }}=\mathrm{f}_{\mathrm{k}} $

$\Rightarrow $ $P+25=37.5$

$ \mathrm{P}=37.5-25=12.5 \mathrm{~N} $

Hence, the correct option is (B).

As the particle falls, two forces act on it :

• Weight $\left(\mathrm{F}_{\mathrm{g}}=\mathrm{mg}\right)$ acting downwards.

• Resistive Force ( $\mathrm{F}=\mathrm{kv}$ ) acting upwards (opposing motion).

The net force on the particle is,

$ \mathrm{F}_{\mathrm{net}}=\mathrm{mg}-\mathrm{kv} $

According to Newton's Second Law, $\mathrm{F}_{\mathrm{net}}=\mathrm{m} \cdot \mathrm{a}=\mathrm{m} \frac{\mathrm{dv}}{\mathrm{dt}}$. Therefore

$ \mathrm{m} \frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{mg}-\mathrm{kv} $

$\Rightarrow $ $ \frac{\mathrm{dv}}{\mathrm{mg}-\mathrm{kv}}=\frac{1}{\mathrm{~m}} \mathrm{dt} $

Integrate both sides, knowing that at $\mathrm{t}=0, \mathrm{v}=0$ (starting from rest) :

$\int\limits_0^{\mathrm{v}} \frac{\mathrm{dv}}{\mathrm{mg}-\mathrm{kv}}=\frac{1}{\mathrm{~m}} \int\limits_0^{\mathrm{t}} \mathrm{dt}$

$\Rightarrow $ $-\frac{1}{\mathrm{k}}[\ln (\mathrm{mg}-\mathrm{kv})]_0^{\mathrm{v}}=\frac{\mathrm{t}}{\mathrm{m}}$

$\Rightarrow $ $\ln \left(\frac{\mathrm{mg}-\mathrm{kv}}{\mathrm{mg}}\right)=-\frac{\mathrm{k}}{\mathrm{m}} \mathrm{t}$

$\Rightarrow $ $\frac{m g-k v}{m g}=e^{-\frac{k}{m} t}$

$\Rightarrow $$1-\frac{\mathrm{kv}}{\mathrm{mg}}=\mathrm{e}^{-\frac{\mathrm{k}}{\mathrm{m}} \mathrm{t}}$

$\Rightarrow $ $ v(t)=\frac{m g}{k}\left(1-e^{-\frac{k}{m} t}\right) $

The resulting equation $\mathrm{v}(\mathrm{t})=\mathrm{v}_{\mathrm{t}}\left(1-\mathrm{e}^{-\frac{\mathrm{t}}{\tau}}\right)$ where $\mathrm{v}_{\mathrm{t}}=\frac{\mathrm{mg}}{\mathrm{k}}$ is the terminal velocity.

At $\mathrm{t}=0, \mathrm{v}=\mathrm{v}_{\mathrm{t}}(1-1)=0$. Hence, the graph must start at the origin $(0,0)$.

As $\mathrm{t} \rightarrow \infty$ the term $\mathrm{e}^{-\frac{\mathrm{k}}{\mathrm{m}} \mathrm{t}}$ approaches 0 , so v approaches a constant value $\frac{\mathrm{mg}}{\mathrm{k}}$ (terminal velocity).

This is an exponential graph with a horizontal asymptote at $\mathrm{v}=\frac{\mathrm{mg}}{\mathrm{k}}$

Hence, the correct option is (D).

Considering the left half of the chain, the mass of this segment is $\frac{m}{2}$.

Weight ( $\mathrm{W}_1$ ) acts vertically downward through the centre of gravity of the half-segment.

$ \mathrm{W}_1=\frac{\mathrm{mg}}{2} $

Tension at support ( $\mathrm{T}_2$ ) acts at an angle of $30^{\circ}$ to the horizontal.

Tension at lowest point $\left(\mathrm{T}_1\right)$ at the lowest point, the chain is perfectly horizontal, so the tension $\mathrm{T}_1$ acts purely in the horizontal direction.

For the half-chain to be in equilibrium, the sum of vertical and horizontal forces must be zero.

The vertical component of the support tension must balance the weight of the half-chain.

$\mathrm{T}_2 \sin \left(30^{\circ}\right)=\frac{\mathrm{mg}}{2}$

$\Rightarrow $ $ \mathrm{T}_2\left(\frac{1}{2}\right)=\frac{\mathrm{mg}}{2} \Rightarrow \mathrm{~T}_2=\mathrm{mg} \ldots (i)$

The horizontal component of the support tension must balance the tension at the lowest point.

$ \mathrm{T}_1=\mathrm{T}_2 \cos \left(30^{\circ}\right) $

From equation (i)

$ \mathrm{T}_1=(\mathrm{mg}) \cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2} \mathrm{mg} $

Therefore, the tension at the lowest point of the chain is $\frac{\sqrt{3}}{2} \mathrm{mg}$.

Hence, the correct option is (B).

When the block is moving up the inclined plane, the component of gravity acting down the slope opposes its motion.

Force along the incline, $\mathrm{F}=-\mathrm{mg} \sin \theta$

Using $\mathrm{F}=$ ma, we get $\mathrm{a}=-\mathrm{g} \sin \theta$.

The initial velocity $\mathrm{v}_{\mathrm{i}}=\mathrm{u}$

The final velocity $\mathrm{v}_{\mathrm{f}}=0$ (since the block comes to rest)

Acceleration $\mathrm{a}=-\mathrm{g} \sin \theta$

using the third equation of motion :

$ \mathrm{v}_{\mathrm{f}}^2=\mathrm{v}_{\mathrm{i}}^2+2 \mathrm{a} S $

Substitute the values into the equation :

$0^2=\mathrm{u}^2+2(-\mathrm{g} \sin \theta) \mathrm{S}$

$\Rightarrow $ $0=\mathrm{u}^2-2 \mathrm{~g} \sin \theta (\mathrm{S})$

$\Rightarrow $ $ S=\frac{u^2}{2 g \sin \theta} $

Hence, the distance travelled is $\frac{u^2}{2 g \sin \theta}$. Hence, correct option is (C).

According to Newton's Second Law $\overrightarrow{\mathrm{F}}=\mathrm{m} \overrightarrow{\mathrm{a}}$.

$ \overrightarrow{\mathrm{a}}=\frac{\overrightarrow{\mathrm{F}}}{\mathrm{~m}} $

Given $\mathrm{m}=4 \mathrm{~kg}$ and $\overrightarrow{\mathrm{F}}=4 \mathrm{t}^3 \hat{\imath}-3 \mathrm{t} \hat{\jmath}$ :

$ \vec{a}=\frac{4 t^3 \hat{\imath}-3 t \hat{\jmath}}{4}=\left(t^3 \hat{\imath}-\frac{3}{4} t \hat{\jmath}\right) \mathrm{m} / \mathrm{s}^2 $

Acceleration is the rate of change of velocity :

$ \vec{a}=\frac{d \vec{v}}{d t} $

Since the mass starts from rest at the origin at $t=0$, we have $\vec{v}(0)=0$

$ \int\limits_0^{\vec{v}} d \vec{v}=\int\limits_0^t\left(t^3 \hat{\imath}-\frac{3}{4} t \hat{\jmath}\right) d t $

$\Rightarrow $ $ \overrightarrow{\mathrm{v}}=\left(\frac{\mathrm{t}^4}{4} \hat{\imath}-\frac{3 \mathrm{t}^2}{8} \hat{\jmath}\right) \mathrm{m} / \mathrm{s} $

So, velocity at $\mathrm{t}=2 \mathrm{~s}$ is :

$ \overrightarrow{\mathrm{v}}=\frac{2^4}{4} \hat{\imath}-\frac{3\left(2^2\right)}{8} \hat{\jmath}=\frac{16}{4} \hat{\imath}-\frac{12}{8} \hat{\jmath} $

$\Rightarrow $ $ \vec{v}=\left(4 \hat{\imath}-\frac{3}{2} \hat{\jmath}\right) \mathrm{m} / \mathrm{s} $

Velocity is the rate of change of position :

$ \vec{v}=\frac{d \vec{r}}{d t} \Rightarrow d \vec{r}=\vec{v} d t $

At $\mathrm{t}=0$, the mass is at the origin ( $\overrightarrow{\mathrm{r}}=0$ ),

$ \int\limits_0^{\overrightarrow{\mathrm{r}}} \mathrm{~d} \overrightarrow{\mathrm{r}}=\int\limits_0^{\mathrm{t}}\left(\frac{\mathrm{t}^4}{4} \hat{\imath}-\frac{3 \mathrm{t}^2}{8} \hat{\jmath}\right) \mathrm{dt} $

$\Rightarrow $ $\overrightarrow{\mathrm{r}}=\left(\frac{\mathrm{t}^5}{20} \hat{\imath}-\frac{\mathrm{t}^3}{8} \hat{\jmath}\right)_0^{\mathrm{t}} \mathrm{m}$

$\Rightarrow $ $\overrightarrow{\mathrm{r}}=\left[\frac{\mathrm{t}^5}{20} \hat{\imath}-\frac{\mathrm{t}^3}{8} \hat{\jmath}\right]-\left[\frac{0^5}{20} \hat{\imath}-\frac{0^3}{8} \hat{\jmath}\right] \mathrm{m}$

$\Rightarrow $ $ \vec{r}=\left(\frac{t^5}{20} \hat{\imath}-\frac{t^3}{8} \hat{\jmath}\right) $

Now, calculate position at $\mathrm{t}=2 \mathrm{~s}$ :

$ \overrightarrow{\mathrm{r}}=\frac{2^5}{20} \hat{\imath}-\frac{2^3}{8} \hat{\jmath}=\frac{32}{20} \hat{\imath}-\frac{8}{8} \hat{\jmath} $

$\Rightarrow $ $\overrightarrow{\mathrm{r}}=\frac{8}{5} \hat{\imath}-1 \hat{\jmath}=\left(\frac{8}{5} \hat{\imath}-\hat{\jmath}\right) \mathrm{m}$

Therefore, at $t=2 \mathrm{~s}$, the velocity is $\vec{v}=4 \hat{\imath}-\frac{3}{2} \hat{\jmath}$ and the position vector is $\vec{r}=\frac{8}{5} \hat{\imath}-\hat{\jmath}$.

Hence, the correct option is (C).

To determine the velocity of the body as it emerges from the force field, we can break this process down as follows:

Force and Acceleration Calculation:

The body is subjected to a constant force of 6 N, which acts along the positive z-axis. The mass of the body is 2 kg.

Using Newton's second law, $ \vec{F} = m \vec{a} $, the acceleration $ \vec{a} $ can be calculated as:

$ \vec{a} = \frac{\vec{F}}{m} = \frac{6 \hat{k}}{2} = 3 \hat{k} \text{ ms}^{-2} $

Initial Velocity:

The initial velocity of the body is given by:

$ \vec{u} = 3 \hat{i} + 4 \hat{j} \text{ ms}^{-1} $

Time of Motion:

The time the body spends in the force field is:

$ t = \frac{5}{3} \text{ seconds} $

Final Velocity Calculation:

The final velocity $ \vec{v} $ is given by the equation of motion:

$ \vec{v} = \vec{u} + \vec{a} t $

Substituting the known values:

$ \vec{v} = (3 \hat{i} + 4 \hat{j}) + 3 \hat{k} \left(\frac{5}{3}\right) $

Simplifying the expression gives:

$ \vec{v} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k} $

Therefore, the velocity of the body as it exits the force field is $ 3 \hat{i} + 4 \hat{j} + 5 \hat{k} \text{ ms}^{-1} $.

To find the force acting on the object, we use the formula $ F = m \times a $, where $ F $ is the force, $ m $ is the mass, and $ a $ is the acceleration.

The velocity $ v $ is given as $ v = 4 \sqrt{x} $.

Squaring both sides, we have $ v^2 = 16x $.

Differentiate $ v^2 $ with respect to $ x $:

$ 2v \frac{dv}{dx} = 16 $

Solving for $ \frac{dv}{dx} $, we find:

$ \frac{v \, dv}{dx} = \frac{16}{2} = 8 $

The force $ F $ is then calculated as:

$ F = 0.5 \times 8 = 4 \, \text{N} $

Thus, the force acting on the object is 4 N.

$\begin{aligned} & m g \sin 60^{\circ}-\mu m g \cos 60^{\circ}=m a \\ & g \sin 60-\mu g \cos 60=\frac{g}{2} \\ & \frac{\sqrt{3}}{2}-\frac{\mu}{2}=\frac{1}{2} \\ & \mu=\sqrt{3}-1 \end{aligned}$

$\begin{aligned} & \mathrm{T}_1 \sin \theta_1+\mathrm{T}_2 \sin \theta_2=\mathrm{mg} \& \mathrm{~T}_1=\sqrt{3} \mathrm{~T}_2 \\ & \Rightarrow \mathrm{~T}_2\left[\sqrt{3} \sin \theta_1+\sin \theta_2\right]=\mathrm{mg} \\ & \text { for } \theta_1=60^{\circ} \& \theta_2=30^{\circ} \\ & \mathrm{T}_2=\frac{\mathrm{mg}}{2} \end{aligned}$

Work Done = Final Kinetic Energy - Initial Kinetic Energy

$W = \Delta K = K_f - K_i$

Calculate the Work Done by the Retarding Force

The problem gives a retarding force $F_r = -kr$. It seems 'r' is a typo for 'x', the position, since the constant $k$ has units of N/m. So, the force is a function of position: $F_r = -kx$.

Because the force isn't constant, we need to find the work done by integrating the force over the distance the block travels in the rough region (from $x = 0.1 \text{ m}$ to $x = 1.9 \text{ m}$).

$W = \int_{0.1}^{1.9} F_r \, dx = \int_{0.1}^{1.9} (-kx) \, dx$

Plugging in $k = 10 \text{ N/m}$:

$W = -10 \int_{0.1}^{1.9} x \, dx$

$W = -10 \left[ \frac{x^2}{2} \right]_{0.1}^{1.9}$

$W = -5 \left[ (1.9)^2 - (0.1)^2 \right]$

$W = -5 [3.61 - 0.01] = -5(3.6)$

$W = -18 \text{ J}$

The negative sign just means the force did negative work, slowing the block down.

Calculate the Initial Kinetic Energy ($K_i$)

The initial kinetic energy is found using the formula $K_i = \frac{1}{2}mv_i^2$.

Given:

Mass $m = 1 \text{ kg}$

Initial speed $v_i = 10 \text{ m/s}$

$K_i = \frac{1}{2}(1 \text{ kg})(10 \text{ m/s})^2 = \frac{1}{2}(100) = 50 \text{ J}$

Use the Work-Energy Theorem to find the Final Speed ($v_f$)

Now we can put everything together into the Work-Energy Theorem equation: $W = K_f - K_i$.

We know $W = -18 \text{ J}$ and $K_i = 50 \text{ J}$. The final kinetic energy is $K_f = \frac{1}{2}mv_f^2$.

$-18 = \left(\frac{1}{2}(1)v_f^2\right) - 50$

Now, let's solve for $v_f$:

$-18 + 50 = \frac{1}{2}v_f^2$

$32 = \frac{1}{2}v_f^2$

$v_f^2 = 64$

$v_f = \sqrt{64}$

$v_f = 8 \text{ m/s}$

The final speed of the block as it leaves the rough region is 8 m/s.

Therefore, the correct option is C.

$\begin{aligned} & \mathrm{T}_1=\mathrm{mg} \cos 30^{\circ} \\ & \mathrm{T}_2=\mathrm{mg} \sin 30^{\circ} \end{aligned}$

$\begin{aligned} &\begin{aligned} & F-m g=m a \\ & F=m a+m g \\ & F-(m-x) g=(m-x) 3 a \end{aligned}\\ &\text { Put F }\\ &\begin{aligned} & \mathrm{Ma}+\mathrm{mg}-\mathrm{mg}+\mathrm{xg}=3 \mathrm{ma}-3 \mathrm{xa} \\ & \mathrm{x}=\frac{2 \mathrm{ma}}{\mathrm{~g}+3 \mathrm{a}} \end{aligned} \end{aligned}$

$\begin{aligned} & T_1 \cos 45=F \\ & \text { and, } T_1 \sin 45=m g \\ & \therefore F=m g \\ & \Rightarrow F=10 \end{aligned}$

$\begin{aligned} & m g \times \frac{L}{2} \cos \theta=N \times L \cos \theta \\ & \Rightarrow \quad N=\frac{m g}{2}=\frac{W}{2} \end{aligned}$

To find the ratio of the masses $\frac{m_2}{m_1}$ given that the acceleration of the system is $\frac{g}{8}$, where $g$ is the acceleration due to gravity, we need to analyze the forces acting on each mass and apply Newton's second law of motion.

For mass $m_1$, the force acting downward (towards the centre of the Earth) is the gravitational force $m_1g$, and the tension $T$ acts upward. Its equation of motion, considering downward as positive, can be given by:

$m_1g - T = m_1a$

For mass $m_2$, the force acting downward is $m_2g$, but since it is on the opposite side of the pulley, the tension $T$ is upwards (considering upward movement as positive direction), so we have:

$T - m_2g = m_2a$

Since the pulley is light and smooth, the tension $T$ is the same on both sides of the pulley. Also, the system accelerates together, so $a = \frac{g}{8}$.

Adding the two equations to eliminate $T$ gives us:

$m_1g - m_2g = (m_1 - m_2)a$

$g(m_1 - m_2) = (m_1 - m_2)\frac{g}{8}$

Canceling out $g$ and dividing both sides by $m_1 - m_2$ (assuming $m_1 \neq m_2$), we get:

$1 = \frac{1}{8}$

This simplification doesn't align with finding the ratio directly, indicating a mistake in handling the simultaneous equations relation to $a$ and $T$. Let's correct our approach to finding the ratio of the masses based on the acceleration and tension.

Since the acceleration $a = \frac{g}{8}$, and considering the system as a whole, the net force causing the acceleration is the difference in gravitational forces on the two masses. This net force provides the system's entire acceleration. Thus, correctly setting up the equations for the system should give:

$m_2g - m_1g = (m_1 + m_2)\frac{g}{8}$

$g(m_2-m_1) = \frac{g}{8}(m_1 + m_2)$

$8(m_2 - m_1) = m_1 + m_2$

$8m_2 - 8m_1 = m_1 + m_2$

$7m_2 = 9m_1$

$\frac{m_2}{m_1} = \frac{9}{7}$

Therefore, the correct ratio of the masses $\frac{m_2}{m_1}$ is $9:7$, which corresponds to Option C.

To determine the coefficient of kinetic friction, let's analyze the time taken by the object to slide down each plane and use the equations of motion for both scenarios.

First, consider the perfectly smooth $45^{\circ}$ inclined plane (no friction). The acceleration of the object on this plane can be calculated using the component of gravitational force parallel to the incline. Since there is no friction, the only force acting down the plane is the component of the gravitational force:

$ a_{\text{smooth}} = g \sin 45^{\circ} = \frac{g}{\sqrt{2}} $

Let the time taken to slide down this smooth plane be $t_{\text{smooth}}$. The distance $d$ covered by the object can be expressed using the equation of motion:

$ d = \frac{1}{2} a_{\text{smooth}} t_{\text{smooth}}^2 = \frac{1}{2} \frac{g}{\sqrt{2}} t_{\text{smooth}}^2 $

Now, consider the rough $45^{\circ}$ inclined plane. The acceleration down the rough plane can be found by considering both the component of gravitational force and the kinetic friction force. Here, the friction force is $f_k = \mu_k N$, where $N = mg \cos 45^\circ = \frac{mg}{\sqrt{2}}$. Thus the frictional force is:

$ f_k = \mu_k \cdot \frac{mg}{\sqrt{2}} $

The net force acting on the object down the plane would be:

$ F_{\text{net}} = mg \sin 45^{\circ} - f_k = \frac{mg}{\sqrt{2}} - \mu_k \cdot \frac{mg}{\sqrt{2}} $

So the net acceleration $a_{\text{rough}}$ is:

$ a_{\text{rough}} = \frac{F_{\text{net}}}{m} = \frac{g}{\sqrt{2}} - \mu_k \cdot \frac{g}{\sqrt{2}} = \frac{g}{\sqrt{2}} (1 - \mu_k) $

The time taken to slide down the rough plane is given as $nt_{\text{smooth}}$. So, the distance $d$ can be written as:

$ d = \frac{1}{2} a_{\text{rough}} (nt_{\text{smooth}})^2 = \frac{1}{2} \left(\frac{g}{\sqrt{2}} (1 - \mu_k)\right) (nt_{\text{smooth}})^2 $

Setting the distances equal for both cases, we get:

$ \frac{1}{2} \frac{g}{\sqrt{2}} t_{\text{smooth}}^2 = \frac{1}{2} \left(\frac{g}{\sqrt{2}} (1 - \mu_k)\right) (nt_{\text{smooth}})^2 $

Simplifying, we find:

$ t_{\text{smooth}}^2 = (1 - \mu_k) n^2 t_{\text{smooth}}^2 $

Therefore:

$ 1 = (1 - \mu_k) n^2 $

Solving for $\mu_k$, we get:

$ \mu_k = 1 - \frac{1}{n^2} $

Thus, the coefficient of kinetic friction between the object and the surface of inclined plane is:

Option A: $1 - \frac{1}{\mathrm{n}^2}$

The force exerted by the ball on the hand can be calculated using the formula derived from Newton's second law of motion, which is $F = \frac{\Delta p}{\Delta t}$, where $F$ is the force, $\Delta p$ represents the change in momentum, and $\Delta t$ is the time over which this change occurs.

The change in momentum, $\Delta p$, can be calculated as the difference between the final momentum, $p_f$, and the initial momentum, $p_i$. In this scenario, because the ball comes to a stop in the player's hand, its final velocity (and hence, its final momentum) is 0. Therefore, the change in momentum is equal to the initial momentum of the ball (since final momentum is zero).

The initial momentum, $p_i$, of the ball can be calculated using the formula $p = mv$, where $m$ is the mass of the ball and $v$ is its velocity. Given that the mass of the ball is $150 \, \mathrm{g} = 0.15 \, \mathrm{kg}$ (converting grams to kilograms) and its velocity is $20 \, \mathrm{m/s}$, we have:

$p_i = (0.15 \, \mathrm{kg}) \times (20 \, \mathrm{m/s}) = 3 \, \mathrm{kg \cdot m/s}$

Since the change in momentum, $\Delta p$, equals the initial momentum ($p_i$) because the final momentum is 0, the force exerted can be found by substituting $\Delta p$ and $\Delta t$ into the first formula:

$F = \frac{3 \, \mathrm{kg \cdot m/s}}{0.1 \, \mathrm{s}} = 30 \, \mathrm{N}$

Therefore, the magnitude of force exerted by the ball on the hand of the player is 30 N, which corresponds to Option C.

To determine the force that the branch pulls on the chain, we need to consider the combined weight of the body and the chain, as this is the total force that the branch must support due to gravity.

The weight of the body is given as $200\, \mathrm{N}$.

To find the weight of the chain, we use the formula for weight, which is the mass of an object multiplied by the acceleration due to gravity ($g$). The mass of the chain is given as $10\, \mathrm{kg}$ and $g = 10\, \mathrm{m/s}^2$, so:

$\text{Weight of the chain} = \text{Mass of the chain} \times g$

$= 10\, \mathrm{kg} \times 10\, \mathrm{m/s}^2$

$= 100\, \mathrm{N}$

The total force that the branch must support is the sum of the weight of the body and the weight of the chain:

$\text{Total force} = \text{Weight of the body} + \text{Weight of the chain}$

$= 200\, \mathrm{N} + 100\, \mathrm{N}$

$= 300\, \mathrm{N}$

Therefore, the branch pulls the chain by a force of $300\, \mathrm{N}$, which corresponds to Option A.

Given that the acceleration of the system is $\frac{g}{\sqrt{2}}$, let's analyze the forces acting on both masses to find the ratio $\frac{m_1}{m_2}$.

The tension force in the string is equal for both masses since the pulley and string are light and smooth. Assuming the pulley only changes the direction of the tension force without affecting its magnitude, we can analyze the forces in the vertical direction for both masses.

The net force acting on $m_1$ (towards the pulley) would be the tension $T$ minus its weight component in the direction of motion, which we will assume is $m_1g$, and for $m_2$, it would be its weight component in the direction of motion, which we can assume is $m_2g$ minus the tension $T$. Given that $m_2$ is moving downwards and $m_1$ is moving upwards, and considering that $m_2>m_1$, the acceleration $a$ of both masses will be the same and can be described as:

For $m_1$:

$T - m_1g = m_1a$

For $m_2$:

$m_2g - T = m_2a$

Adding these equations to eliminate $T$ gives:

$m_2g - m_1g = m_1a + m_2a$

Given that the acceleration $a$ of the system is $\frac{g}{\sqrt{2}}$, we can substitute $a$ with $\frac{g}{\sqrt{2}}$ in the equation:

$m_2g - m_1g = m_1\left(\frac{g}{\sqrt{2}}\right) + m_2\left(\frac{g}{\sqrt{2}}\right)$

Simplifying and factoring out $g$, we get:

$m_2 - m_1 = \frac{m_1}{\sqrt{2}} + \frac{m_2}{\sqrt{2}}$

Multiplying every term by $\sqrt{2}$ to clear the denominator, we get:

$\sqrt{2}(m_2 - m_1) = m_1 + m_2$

Rearranging the terms to isolate the masses on one side, we obtain:

$\sqrt{2}m_2 - m_2 = m_1(\sqrt{2} + 1)$

This simplifies to:

$m_2(\sqrt{2} - 1) = m_1(\sqrt{2} + 1)$

Hence, the ratio $\frac{m_1}{m_2}$ is:

$\frac{m_1}{m_2} = \frac{\sqrt{2} - 1}{\sqrt{2} + 1}$

This corresponds to Option C $\frac{\sqrt{2}-1}{\sqrt{2}+1}$.

To find the angle between $\vec{F}$ and $\overrightarrow{\mathrm{p}}$, we first need to understand the relationship between force and momentum. The force $\vec{F}$ acting on a particle is related to the rate of change of its linear momentum $\overrightarrow{\mathrm{p}}$ with respect to time, as described by Newton's second law of motion:

$\vec{F} = \frac{d\overrightarrow{\mathrm{p}}}{dt}$

Given the expression for the momentum $\overrightarrow{\mathrm{p}}(t) = \hat{i} \cos (kt) - \hat{j} \sin (kt)$, we can find $\vec{F}$ by differentiating $\overrightarrow{\mathrm{p}}$ with respect to $t$:

$\frac{d\overrightarrow{\mathrm{p}}}{dt} = -k\hat{i} \sin (kt) - k\hat{j} \cos (kt)$

So, $\vec{F} = -k\hat{i} \sin (kt) - k\hat{j} \cos (kt)$.

Now, to find the angle between $\vec{F}$ and $\overrightarrow{\mathrm{p}}$, we use the dot product formula:

$\vec{F} \cdot \overrightarrow{\mathrm{p}} = |\vec{F}| |\overrightarrow{\mathrm{p}}| \cos(\theta)$,

where $\theta$ is the angle between $\vec{F}$ and $\overrightarrow{\mathrm{p}}$. However, in this case, it's more insightful to see if $\vec{F}$ and $\overrightarrow{\mathrm{p}}$ are orthogonal (at a $\frac{\pi}{2}$ angle to each other), because the dot product of two perpendicular vectors is zero.

The dot product of $\vec{F}$ and $\overrightarrow{\mathrm{p}}$ is:

$(-k\hat{i} \sin (kt) - k\hat{j} \cos (kt)) \cdot (\hat{i} \cos (kt) - \hat{j} \sin (kt)) =$

$- k \sin (kt) \cos (kt) + k \cos (kt) \sin (kt) = 0$

The result is zero, indicating that the angle between $\vec{F}$ and $\overrightarrow{\mathrm{p}}$ is indeed $\frac{\pi}{2}$.

Therefore, the correct option is:

Option A $\frac{\pi}{2}$.

To find the force of kinetic friction acting on the box, we can use the formula for kinetic friction, which is given by:

$F_{\text{friction}} = \mu_{k} \cdot N$

where:

• $F_{\text{friction}}$ is the force of friction,
• $\mu_{k}$ is the coefficient of kinetic friction,
• $N$ is the normal force exerted by the surface onto the box.

Since the box is moving on a horizontal surface, the normal force $N$ would be equal to the weight of the box, which is calculated by $mg$, where $m$ is the mass of the box and $g$ is the acceleration due to gravity. For most calculations, $g$ is approximated as $9.8 \, \text{m/s}^2$.

Substituting the given values:

• $m = 50 \, \text{kg}$,
• $\mu_k = 0.3$,
• $g = 9.8 \, \text{m/s}^2$.

First, calculate the weight of the box, which is the normal force:

$N = mg = 50 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 490 \, \text{N}$

Then, use this value to find the force of kinetic friction:

$F_{\text{friction}} = \mu_{k} \cdot N = 0.3 \times 490 \, \text{N} = 147 \, \text{N}$

Thus, the force of kinetic friction acting on the box is 147 N. The correct answer is Option B.

To solve this problem, we'll analyze the forces acting on the combined system of the wooden block and the iron cylinder as they accelerate downward due to the yielding floor.

Given:

Mass of wooden block, $ m_1 = 5 \, \text{kg} $

Mass of iron cylinder, $ m_2 = 25 \, \text{kg} $

Total mass of the system, $ m = m_1 + m_2 = 30 \, \text{kg} $

Acceleration of the system, $ a = 0.1 \, \text{m/s}^2 $ (downward)

Acceleration due to gravity, $ g = 9.8 \, \text{m/s}^2 $

Step 1: Draw the Free Body Diagram (FBD)

For the combined system:

Downward Forces:

Weight of the system: $ W = m \times g = 30 \times 9.8 = 294 \, \text{N} $

Upward Forces:

Normal force from the floor: $ N $

Step 2: Apply Newton's Second Law

Since the system is accelerating downward, the net force is:

$ \text{Net Force} = \text{Total Downward Force} - \text{Total Upward Force} $

According to Newton's second law:

$ m \times a = W - N $

Step 3: Solve for the Normal Force ($ N $)

$ N = W - m \times a $

Substitute the given values:

$ N = 294 \, \text{N} - 30 \, \text{kg} \times 0.1 \, \text{m/s}^2 $

$ N = 294 \, \text{N} - 3 \, \text{N} $

$ N = 291 \, \text{N} $

Step 4: Interpret the Result

The normal force ($ N $) is the force exerted by the floor on the system upward. By Newton's third law, the action force of the system on the floor is equal in magnitude and opposite in direction to the normal force.

Therefore, the action force of the system on the floor is $ 291 \, \text{N} $.

$\begin{aligned} & m g \sin \theta=\mu \cdot m g \cos \theta \\ & \Rightarrow \tan \theta=\mu=1 \end{aligned}$

Acceleration (a) $=\frac{f-F}{m}$

When applied force became equal to $f_{\max }$, block will start moving.

As $F$ increases linearly, so acceleration of also moving block will increase linearly.

To find the acceleration acting on the body, we first need to determine the resultant force acting on the body by adding the two forces $\vec{F}_1$ and $\vec{F}_2$ vectorially. Then, we apply Newton's second law of motion, which states that the acceleration $\vec{a}$ of a body is directly proportional to the total force $\vec{F}$ acting on it and inversely proportional to the mass $m$ of the body :

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

or

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

Let's start by adding the forces:

$ \vec{F}_1 + \vec{F}_2 = (5 \hat{i}+8 \hat{j}+7 \hat{k}) + (3 \hat{i}-4 \hat{j}-3 \hat{k}) $

Performing the addition component-wise:

$ \vec{F}_{\text{total}} = (5 + 3)\hat{i} + (8 - 4)\hat{j} + (7 - 3)\hat{k} \ \vec{F}_{\text{total}} = 8 \hat{i} + 4 \hat{j} + 4 \hat{k} $

Now, let's use the formula for acceleration with $m = 4 \mathrm{~kg}$:

$ \vec{a} = \frac{\vec{F}_{\text{total}}}{m} = \frac{8 \hat{i} + 4 \hat{j} + 4 \hat{k}}{4 \mathrm{~kg}} $

Divide each component by the mass:

$ \vec{a} = 2 \hat{i} + 1 \hat{j} + 1 \hat{k} $

So, the acceleration acting on the body is:

$ \vec{a} = 2 \hat{i} + \hat{j} + \hat{k}$

Thus, the correct option is:

Option A :

$2 \hat{i}+\hat{j}+\hat{k}$

The first step in solving this problem is to calculate the change in momentum of the ball when it is caught. The change in momentum, or impulse, is the product of the mass of the ball and the change in velocity (as momentum is mass times velocity).

The ball is initially moving with a velocity of $v_i = 25 \mathrm{~m/s}$ before the catch and finally comes to rest with a velocity of $v_f = 0 \mathrm{~m/s}$ after the catch. Since the ball is caught, the final velocity is zero. The change in velocity $\Delta v = v_f - v_i = 0 - 25 = -25 \mathrm{~m/s}.$ Remember that the direction of the force exerted by the ball on the hand will be opposite to the direction of the ball's initial motion.

The mass of the ball $m$ is given as $120 \mathrm{~g}$ which needs to be converted into kilograms to maintain SI units: $m = 120 \mathrm{~g} = 120 \times 10^{-3} \mathrm{~kg} = 0.12 \mathrm{~kg}.$

Now we can calculate the change in momentum (impulse): $\Delta p = m \Delta v = 0.12 \mathrm{~kg} \times (-25 \mathrm{~m/s}).$

Substituting the values we get: $\Delta p = 0.12 \times -25 = -3 \mathrm{~kg \cdot m/s}.$

The negative sign indicates that the change in momentum is in the opposite direction of the ball's initial motion, which makes sense because the ball's velocity is reduced to zero.

The magnitude of the impulse is independent of the sign and is $3 \mathrm{~kg \cdot m/s}$.

Impulse is also equal to the average force exerted on the ball times the time interval during which the force is exerted. We can use the formula: $\Delta p = F_{avg} \Delta t$

Where $F_{avg}$ is the average force and $\Delta t$ is the time interval of $0.1 \mathrm{~s}$. Re-arranging the formula to solve for $F_{avg}$ gives us: $F_{avg} = \frac{\Delta p}{\Delta t}.$

Substituting the known values we have: $F_{avg} = \frac{3}{0.1} = 30 \mathrm{~N}.$

The magnitude of the average force exerted by the hand of the player to catch the ball is $30 \mathrm{~N}$.

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

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

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

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

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.

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