Laws of Motion

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

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

Impulse of a force = Area between the force-time graph

Hence, impulse of force = area of $\Delta$OA'A + area of rectangle A'B'BA + area of trapezium BB'C'C + area of rectangle CC'ED

$ = {1 \over 2} \times 1.5 \times 3 + 1 \times 3 + {1 \over 2}(3 + 2)(3 - 2.5) + 2 \times 1$

$ = 2.25 + 3 + 1.25 + 2$

$ = 8.50$ Ns

According to the question,

Let OB = x, OA = y and x = L cos$\theta$

So, by Pythagoras theorem,

$\Rightarrow$ x² + y² = L²

Differentiating w.r.t. 't', we get

$2x{{dx} \over {dt}} + 2y{{dy} \over {dt}} = 0$

$2x{v_B} + 2y{v_A} = 0$ [$\because$ ${{dx} \over {dt}} = {v_B}$ and ${{dy} \over {dt}} = {v_A}$]

$ \Rightarrow {v_A} = - {x \over y}{v_B}$

$ \Rightarrow \left| {{v_A}} \right| = {x \over y}{v_B} = {v_B}\cot \theta $ [$\because$ ${x \over y} = \cot \theta $]

$ = \sqrt 3 \times \cot 30^\circ $

$ = \sqrt 3 \times {{\sqrt 3 }}$ = 3 m/s

To find the friction force on the block, we first need to analyze the forces acting on the block on an inclined plane.

The forces acting on the block include:

• Gravitational force (weight) acting vertically downward, $mg$
• Normal force, $N$, exerted by the plane perpendicular to its surface
• Frictional force, which acts parallel to the inclined surface and opposite to the direction of possible motion

The gravitational force can be resolved into two components:

• Perpendicular to the inclined plane: $mg \cos \theta$
• Parallel to the inclined plane: $mg \sin \theta$

In this problem:

• Mass of the block, $m$ = 10 kg
• Angle of the incline, $\theta$ = 30$^\circ$
• Coefficient of static friction, $\mu_s$ = 0.1

First, we calculate the perpendicular component of the gravitational force to find the normal force:

$N = mg \cos \theta$

Substituting the values:

$N = (10 \, \text{kg}) (9.8 \, \text{m/s}^2) \cos 30^\circ$

$\cos 30^\circ = \frac{\sqrt{3}}{2}$

$N = 10 \times 9.8 \times \frac{\sqrt{3}}{2}$

$N = 49 \sqrt{3} \, \text{N}$

Next, we find the maximum static friction force:

$f_{\text{max}} = \mu_s N$

$f_{\text{max}} = 0.1 \times 49 \sqrt{3} \, \text{N}$

$f_{\text{max}} = 4.9 \sqrt{3} \, \text{N}$

We now compare this with the component of the gravitational force parallel to the inclined plane:

$mg \sin \theta$

Substituting the values:

$mg \sin 30^\circ = 10 \times 9.8 \times \frac{1}{2}$

$mg \sin 30^\circ = 49 \, \text{N}$

The friction force will be equal to the component of the gravitational force parallel to the inclined plane but within the limit of the maximum static friction force. Here, we recognize that the maximum static friction force (4.9 $\sqrt{3}$ N) is indeed greater than 4.9 N (the component of the gravitational force parallel to the incline). Hence, the actual friction force counteracting the motion will be:

$49 \, \text{N}$

Therefore, the correct answer is Option C:

Option C: 49 N

Given, mass of object, $m_2=2 \mathrm{~kg}$ Velocity, $\mathbf{v}=\left(8 t \hat{\mathbf{i}}+3 t^2 \hat{\mathbf{j}}\right) \mathrm{m} / \mathrm{s}$

∴ Acceleration of object is given as

$ \begin{aligned} & \mathbf{a}=\frac{d \mathbf{v}}{d t}=\frac{d}{d t}\left(8 t \hat{\mathbf{i}}+3 t^2 \hat{\mathbf{j}}\right) \\ & \mathbf{a}=8 \hat{\mathbf{i}}+6 t \hat{\mathbf{j}} \mathrm{~ms}^{-2} \end{aligned} $

Hence, according to Newton's second law of motion, force on the object,

$ \Rightarrow \quad \begin{align*} \mathbf{F} & =m \mathbf{a}=2(8 \hat{\mathbf{i}}+6 t \hat{\mathbf{j}}) \\ \mathbf{F} & =16 \hat{\mathbf{i}}+12 t \hat{\mathbf{j}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ |\mathbf{F}| & =\sqrt{16^2+(12 t)^2} \end{align*} $

$ \begin{array}{lc} \Rightarrow & 20=\sqrt{16^2+(12 t)^2} \\ & {[\text { Given, }|\mathbf{F}|=20 \mathrm{~N}]} \\ \Rightarrow & 400=256+144 t^2 \\ \Rightarrow & 144 t^2=144 \Rightarrow t^2=1 \Rightarrow t=1 \mathrm{~s} \end{array} $

Put $t=1 \mathrm{~s}$ in Eq. (i), we get

$ \mathbf{F}=16 \hat{\mathbf{i}}+12 \hat{\mathbf{j}} $

∴ Angle made by force $\mathbf{F}$ relative to positive direction of $X$-axis is given as

$ \begin{aligned} \theta & =\tan ^{-1}\left(\frac{F_Y}{F_X}\right) \\ & =\tan ^{-1}\left(\frac{12}{16}\right)=\tan ^{-1}\left(\frac{3}{4}\right) \end{aligned} $

All forces are resolved in two perpendicular axes ( $X$ and $Y$ ) as shown in the figure.

Since, block of mass $m$ is in equilibrium.

Hence, resolving the forces in $x$-direction, we get

$ \begin{aligned} \left|\mathbf{F}_2\right| \cos \left(60^{\circ}\right)= & \left|\mathbf{F}_1\right| \cos \left(30^{\circ}\right) \\ \left|\mathbf{F}_2\right| \times \frac{1}{2}= & 10 \times \frac{\sqrt{3}}{2} \\ & \left(\because\left|\mathbf{F}_1\right|=10 \mathrm{~N} \text { given }\right) \end{aligned} $

$ \left|\mathrm{F}_2\right|=10 \sqrt{3} \mathrm{~N} $

Again, resolving the force in $y$-direction, we get

$ \begin{aligned} \left|\mathbf{F}_3\right| & =\left|\mathbf{F}_1\right| \sin \left(30^{\circ}\right)+\left|\mathbf{F}_2\right| \sin \left(60^{\circ}\right) \\ & =10 \times \frac{1}{2}+10 \sqrt{3} \times \frac{\sqrt{3}}{2} \\ & =5+15=20 \mathrm{~N} \end{aligned} $

The given situation is shown below

Normal reaction is

$ \mathbf{N}=F \cos \theta=F \cos 30^{\circ} $

Friction force when block is just on the verge of sliding is

$ f=\mu N=\mu F \cos 30^{\circ}=\frac{3}{2} F[\therefore \mu=\sqrt{3}] $

As block is in equilibrium, friction $=$ weight of block $(m g)+$ vertical component of force ( $F \sin \theta$ )

$ \Rightarrow \quad \frac{3}{2} F=3 \times 10+F \times \frac{1}{2} \Rightarrow F=30 \mathrm{~N} $

Given,

Momentum of bullet after firing = 20 $\mathrm{kg} \mathrm{ms}^{-1}$

Velocity $=1000 \mathrm{~ms}^{-1}$

We know that, $p=m v$

$ \begin{aligned} & \Rightarrow m=\frac{p}{v}=\frac{20}{1000} \mathrm{~kg} \\ & \Rightarrow m=\frac{20}{1000} \times 1000 \mathrm{~g} \Rightarrow m=20 \mathrm{~g} \\ & \Rightarrow \quad 20 \leq \frac{u^2 \sin 2 \theta}{g} \leq 24.2 \\ & \Rightarrow \quad 20 \leq \frac{u^2 \times \sin \left(2 \times 15^{\circ}\right)}{10} \leq 24.2 \\ & \Rightarrow \quad 200 \leq u^2 \times \frac{1}{2} \leq 242 \quad\left(\because \sin 30^{\circ}=\frac{1}{2}\right) \\ & \Rightarrow \quad 400 \leq u^2 \leq 484 \\ & \Rightarrow \quad 20 \mathrm{~m} / \mathrm{s} \leq u \leq 22 \mathrm{~m} / \mathrm{s} \end{aligned} $

So, option (a) is best choice.

The given situation is shown in the figure. Resolving applied force, we have following forces on block

$ \text { Given, } \tan \theta=\frac{3}{4}, \sin \theta=\frac{3}{5} \text { and } \cos \theta=\frac{4}{5} $

So, horizontal component,

$ F_1=12 \times \cos \theta=12 \times \frac{4}{5}=\frac{48}{5} \mathrm{~N} $

and vertical component,

$ F_2=12 \times \sin \theta=12 \times \frac{3}{5}=\frac{36}{5} \mathrm{~N} $

So, total applied force on ground is

$ 10+\frac{36}{5}+2 \times 10=37.2 \mathrm{~N} $

So, reaction $N_1$ of ground $=37.2 \mathrm{~N}$ upwards

Also, total force horizontally on wall is

$ F_1=\frac{48}{5}=9.6 \mathrm{~N} $

So, reaction $N_2$ of wall $=9.6 \mathrm{~N}$ towards left

Initial momentum, P_i = mv

${P_i} = m\sqrt {2gh} $ [$\because$ $v\sqrt {2gh} $]

New height of body, ${h_1} = h + 3h = 4h$

Now, final momentum,

${P_f} = m\sqrt {2g\,.\,{h_1}} $

$ = m\sqrt {2g(4h)} $

$ = m\,.\,2\sqrt {2gh} = 2{P_i}$

$\therefore$ % change in momentum = ${{{P_f} - {P_i}} \over {{P_i}}} \times 100$

$ = {{2{P_i} - {P_i}} \over {{P_i}}} \times 100$

$ = 100\% $

Given, mass of plane, M = 10000 kg

Coefficient of friction, $\mu$ = 0.2

Distance, s = 50 m

Initial speed, u = 0

Final speed, v = 72 kmh^$-$1 = 20 ms^$-$1

Let a be the acceleration of the plane.

From third equation of motion,

v² $-$ u² = 2as

(20)² $-$ (0)² = 2 $\times$ a $\times$ 50

$\therefore$ $a = {{20 \times 20} \over {100}} = 4$ ms^$-$2

Force required to provide necessary acceleration to the plane,

F₁ = ma = 10000 $\times$ 4 = 4 $\times$ 10⁴ N

Force required to overcome friction by engine,

F₂ = $\mu$R = $\mu$mg = 0.2 $\times$ 10000 $\times$ 9.8

= 1.96 $\times$ 10⁴ N

Therefore, the minimum force required by the engine,

F₁ + F₂ = 4 $\times$ 10⁴ + 1.96 $\times$ 10⁴ = 5.96 $\times$ 10⁴ N