Laws of Motion

Breaking strength of string is $\frac{4}{3}$ times weight of person

So, maximum tension with stand by string

$ T=\frac{4}{3} W=\frac{4}{3} m g $

Now, for maximum acceleration while climbing up;

$ \begin{aligned} & T-m g=m a \\ & \frac{4}{3} m g-m g=m a \\ & a=\frac{4}{3} g-g=\frac{g}{3} \end{aligned} $

$ \begin{aligned} &\text { } F-f_k=m a\\ &\begin{array}{ll} \Rightarrow & 20-f_k=2 \times 7 \\ \Rightarrow & f_k=6 \Rightarrow \mu_k N=6 \\ \Rightarrow & \mu_k=\frac{6}{N}=\frac{6}{m g}=\frac{6}{2 \times 10}=0.3 \end{array} \end{aligned} $

$ \begin{aligned} &\text { Apparent weight when lift moves up }\\ &=m(g+a)=30(10+2)=360 \mathrm{~N} \end{aligned} $

Let $S=$ displacement of body beforce coming to rest

Then using,

$ v^2-u^2=2 a S $

we get,

$ \begin{aligned} -u^2 & =2\left(\frac{-F}{m}\right) S \\or\,\, S & =\frac{u^2 m}{2 F} \text { or } S \propto u^2 m \end{aligned} $

(for same retarding force $F$ )

$ \therefore \quad \frac{S_A}{S_B}=\frac{u_A^2 m_A}{u_B^2 m_B}=\frac{400 \times 1.5}{225 \times 3}=\frac{8}{9} $

Step 1: Find the friction force for each block

The friction force is given by: $ F_{friction} = \mu_k \times m \times g $

For block $A$ (mass $=2\, \mathrm{kg}$):
$ F_{K A} = 0.3 \times 2 \times 10 = 6\,\mathrm{N} $

For block $B$ (mass $=4\, \mathrm{kg}$):
$ F_{K B} = 0.3 \times 4 \times 10 = 12\,\mathrm{N} $

Step 2: Find the net force on each block

The net force is the applied force minus friction:

For block $A$:
$ F_A = 20 - 6 = 14\,\mathrm{N} $

For block $B$:
$ F_B = 20 - 12 = 8\,\mathrm{N} $

Step 3: Find the acceleration of each block

Use $a = \frac{F_{net}}{m}$ for each block.

For block $A$:
$ a_A = \frac{14}{2} = 7\,\mathrm{m/s}^2 $

For block $B$:
$ a_B = \frac{8}{4} = 2\,\mathrm{m/s}^2 $

Step 4: Find the ratio of accelerations

$ \frac{a_A}{a_B} = \frac{7}{2} $ So, the ratio is $7:2$.

$ \begin{aligned} &\text { Using Lami's theorem }\\ &\frac{T}{\sin 90^{\circ}}=\frac{30}{\sin \left(180^{\circ}-30^{\circ}\right)}=\frac{W}{\sin \left(90^{\circ}+30^{\circ}\right)} \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad T=\frac{30}{\sin 30^{\circ}}=\frac{W}{\cos 30^{\circ}} \\ & \Rightarrow \quad T=60=\frac{2 W}{\sqrt{3}} \\ & \therefore \quad T=60 \mathrm{~N} \\ & \quad \frac{2 W}{\sqrt{3}}=60 \Rightarrow W=30 \sqrt{3} \mathrm{~N} \end{aligned} $

When the balloon is descending, the net force acting on it is given by $F_{\text {net }}=m g-B=m a$, where $B$ is the Buoyant force.

Thus,  $ B=m g-m a \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

For the balloon to move upwards with the same acceleration $a$, a mass $\Delta m$ must be removed. The new net force equation becomes

$ \begin{aligned} B-(m-\Delta m) g & =(m-\Delta m) a \\ m g-m a-(m-\Delta m) g & =(m-\Delta m) a\,\,\,\, [From Eq. (i)]\end{aligned} $

$ \Rightarrow \quad \Delta m=\frac{2 m a}{g+a} $

Frictional force

$ \begin{aligned} & f_s=\mu m g=0.2 \times 10 \times 10 \\ \Rightarrow \quad & f_s=20 \mathrm{~N} \end{aligned} $

$\therefore$ deceleration, $a=\frac{f_s}{m}$

$ =\frac{20}{10}=2 \mathrm{~m} / \mathrm{s}^2 $

Using, $v^2=u^2+2 a s$

$ \begin{aligned} \Rightarrow \quad s & =\frac{v^2-u^2}{2 a} \\ & =\frac{0^2-2^2}{2 \times(-2)}=1 \mathrm{~m} \end{aligned} $

Maximum static friction

$ \left(f_s\right)_{\max }=\mu_s N $

For the block not to move,

$ \begin{aligned} & F \cos 60^{\circ}=\left(f_s\right)_{\max } \\ & F \cos 60^{\circ}=\mu_s N \end{aligned} $

$ \begin{aligned} & F \cos 60^{\circ}=\mu_s\left(m g+F \sin 60^{\circ}\right) \\ & \Rightarrow F \times \frac{1}{2}=\frac{1}{2 \sqrt{3}}\left(\sqrt{3} \times 10+F \frac{\sqrt{3}}{2}\right) \\ & \Rightarrow \quad F=\frac{1}{\sqrt{3}}\left(10 \sqrt{3}+\frac{F \sqrt{3}}{2}\right) \\ & \Rightarrow \quad F=10+\frac{F}{2} \\ & \Rightarrow \quad F-\frac{F}{2}=10 \\ & \quad F=20 \mathrm{~N} \end{aligned} $

$ \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} $

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

Initial speed of body on the horizontal rough surfaces, $u=10 \mathrm{~ms}^{-1}$

Final velocity, $v=4 \mathrm{~ms}^{-1}$ and $t=2 \mathrm{~s}$.

If $a$ be the retardation due to friction, then by using equation,

$\begin{aligned} & v=u-a t \\ & \Rightarrow \quad 4=10-a \times 2 \\ \end{aligned}$

$\Rightarrow \quad 2 a=6 \Rightarrow a=3 \mathrm{~ms}^{-2}$

According to given situation,

$\begin{array}{ll} \text { friction force }=m a \\ \Rightarrow \mu_k m g=m a \\ \Rightarrow \mu_k=\frac{a}{g}=\frac{3}{10}=0.3 \end{array}$

Mass of cricket ball, $m=50 \mathrm{~g}=0.05 \mathrm{~kg}$

Initial velocity of ball, $u=50 \mathrm{cms}^{-1}=0.5 \mathrm{~ms}^{-1}$

time, $t=0.5 \mathrm{~s}$

If $a$ be the retardation applied on the cricket ball, then by using

$\begin{aligned} v & =u-a t \\ \Rightarrow \quad 0 & =0.5-a \times 0.5 \Rightarrow a=\frac{0.5}{0.5} \\ & =1 \mathrm{~m} / \mathrm{s}^2 \end{aligned}$

Force applied,

$\begin{aligned} F & =\text { mass } \times \text { acceleration } \\ & =0.05 \times 1=0.05 \mathrm{~N} \end{aligned}$

FBD of the situation is given below.

From Newton's law, $\Sigma \mathbf{F}=$ ma

Force on $M_1$,

in $x$-direction, $T=M_1 a$ ... (i)

in $y$-direction, $N_1-w_1=0$ .... (ii)

Similarly, forces on $M_2$,

in $x$-direction, $M_2 g \sin \theta-T=M_2 a$ .... (iii)

in $y$-direction, $N_2-m_2 g \cos \theta=0$ .... (iv)

Adding to eq. (i) with eq. (ii), we get

$\begin{aligned} M_2 g \sin \theta & =M_1 a+M_2 a \\ \Rightarrow \quad a & =\frac{M_2 g \sin \theta}{M_1+M_2} \end{aligned}$

According to question, when $M_1$ is doubled and $M_2$ is halved

$a^{\prime}=\frac{\frac{M_2}{2} g \sin \theta}{2 M_1+\frac{M_2}{2}}=\frac{M_2 g \sin \theta}{4 M_1+M_2}$

Now, $\frac{a^{\prime}}{a}=\frac{\frac{M_2 g \sin \theta}{4 M_1+M_2}}{\frac{M_2 g \sin \theta}{M_1+M_2}} \Rightarrow a^{\prime}=\left(\frac{M_1+M_2}{4 M_1+M_2}\right) a$

The given situation is shown below

Let $a$ be the common acceleration in the system of combination of blocks, then equation of motion of block $m_1$ is

$\begin{aligned} & T=m_1 a \\ & T=40 a \quad \text{.... (i)} \end{aligned}$

and equation of motion of block $m_2$ is

$\begin{aligned} F-T & =m_2 a \\ \Rightarrow 1000-T & =m_2 a \\ \Rightarrow 1000-T & =60 \times\left(\frac{T}{40}\right) \quad \text{[from Eq. (i)]}\\ \Rightarrow 1000-T & =\frac{3}{2} T \Rightarrow 2000-2 T=3 T \\ \Rightarrow \quad & T=\frac{2000}{5}=400 \mathrm{~N} \end{aligned}$

Given, mass of bock $A, m_A=10 \mathrm{~kg}$

Mass of slab $B, m_B=30 \mathrm{~kg}$

Coefficient of static $\left(\mu_s\right)$ and kinetic friction, $\left(\mu_k\right)=0.6$ and 0.4

Applied force, $F=100 \mathrm{~N}$

Acceleration due to gravity, $g=10 \mathrm{~ms}^{-2}$

According to given figure,

$N_B$ is normal reaction on slab, $B=40 \mathrm{~g}$

$N_A$ is normal reaction on block, $A=10 \mathrm{~g}$

Image

Static friction force $\left(f_s\right)=\mu_s N_A$

$=\frac{6}{10} \times 100=60 \mathrm{~N}$

Kinetic friction force $\left(f_k\right)=\mu_k N_A$

$=\frac{4}{10} \times 100=40 \mathrm{~N}$

$\because \quad F > f_s$

$\therefore$ There will be relative motion and force on slab $B, F_B=f_k=40 \mathrm{~N}$

Acceleration of slab, $B=\frac{40}{30}=1.33 \mathrm{~ms}^{-2}$

$\simeq 1.31 \mathrm{~ms}^{-2}$

As we know that,

W is weight of body and N be the normal reaction.

Therefore, angle between W and N = 180$\Upsilon$.

Given, mass of block A and B are $m_A=4$ kg and $m_B=6$ kg

Friction between A and B, $f=12$ N

Acceleration due to gravity, $g=10$ ms$^{-2}$

In order to move complete system together,

$F_B=(m_A+m_B)a$

$=(4+6)a$

$=10a$ .... (i)

and $F_A=12=4a\Rightarrow a=3$

Substituting in Eq. (i), we get

$F_B=10\times3=30$ N

According to the question,

Time taken by ball to reach on floor when lift is stationary is t$_1$ and time taken by ball to reach on floor when lift is moving up with constant

Since, $s=ut+1/2\,at^2$

Therefore, in case 2

$s=0.t_1+1/2\,gt^2$ [$\because a=g$]

$\Rightarrow s=\frac{1}{2}gt_1^2$ ..... (i)

If $a_\mathrm{net}$ is net acceleration of ball, then in case 2 when lift is going up with acceleration a

$a_\mathrm{net}=(g+a)$

$s=\frac{1}{2}(g+a)t_2^2$ .... (ii)

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

$1/2gt_1^2=1/2(g+a)t_2^2$

$\therefore \quad t_1 > t_2$

Given, coefficient of friction = $\propto$

Angle of inclination $=\theta$

Acceleration due to gravity $=g$

Now, from the free body diagram of inclined plane and mass system.

Force along the plane, $m g \sin \theta-f=m a$

Here,

$f =\propto N$

$\Rightarrow \quad m g \sin \theta-\propto N =m a$ ..... (i)

and force along line perpendicular to plane will be

$N=m g \cos \theta$ ..... (ii)

where, $m, f, a$ and $N$ are mass, friction force, acceleration and normal reaction experienced by the body.

Now, from Eqs. (i) and (ii), we get

$\begin{aligned} m g \sin \theta-\propto m g \cos \theta & =m a \\ \Rightarrow \quad a & =g(\sin \theta-\propto \cos \theta) \end{aligned}$