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

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

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

Let v be speed of the bead when it makes an angle $\theta$ with the vertical direction.

The forces acting on the bead are its weight mg and normal reaction N. Since all the forces acting on the bead are conservative, the mechanical energy of the bead is conserved i.e.,

$mgr(1 - \cos \theta ) = {1 \over 2}m{v^2}$ ..... (1)

The radially inward components of force provide centripetal acceleration to the bead i.e.,

$mg\cos \theta - N = m{v^2}/r$ ...... (2)

Eliminate v² from equations (1) and (2) to get

$N = mg(3\cos \theta - 2)$ ...... (3)

When 3cos$\theta$ > 2, direction of N will be same as shown in figure, i.e., radially outward.

When 3cos$\theta$ < 2, direction of N will be radially inward. Now, normal on the bead is opposite to the force applied on the wire by the bead, following Newton's third law. Therefore, the force applied on the bead by the wire is radially inwards initially and radially outwards later, as it moves from A to B.

The forces on the block of mass m₁ are its weight m₁g, normal reaction from the inclined plane N₁, and the reaction from the second block R.

Similarly, forces on the block of mass m₂ are m₂g, N₂, R, and the frictional force f. If $\theta$ is slowly increased, f starts increasing and attains its maximum value f = $\mu$N₂ at $\theta$ = $\theta$_r (angle of repose). The blocks are stationary if $\theta$ $\le$ $\theta$_r otherwise they are moving. Consider the limiting case, $\theta$ = $\theta$_r, when the blocks are at rest and

$f = \mu {N_2}$ ...... (1)

Apply Newton's second law to m₁,

$R = {m_1}g\sin \theta $ ....... (2)

${N_1} = R = {m_2}g\cos \theta $ ....... (3)

and to m₂,

$f = R + {m_2}g\sin \theta $ ...... (4)

${N_2} = {m_1}g\cos \theta $ ...... (5)

Eliminate R, N₂ and f from equations (1)-(5) to get

${\theta _r} = {\tan ^{ - 1}}\left( {{{\mu {m_2}} \over {{m_1} + {m_2}}}} \right) = {\tan ^{ - 1}}\left( {{{0.3 \times 2} \over {1 + 2}}} \right)$

$ = {\tan ^{ - 1}}(0.2) = 11.5^\circ $

Thus, for $\theta$ = 5$^\circ$ and $\theta$ = 10$^\circ$, blocks are at rest with frictional force

$f = R + {m_2}g\sin \theta = ({m_1} + {m_2})g\sin \theta $

For $\theta$ = 15$^\circ$ and $\theta$ = 20$^\circ$, blocks are moving with frictional force

$f = \mu {N_2} = \mu {m_2}g\cos \theta $, (limiting value).

The forces acting on the block are F = 1 N towards the left, weight mg = 0.1 $\times$ 10 = 1 N downwards, normal force N, and the frictional force f. Resolve F and mg along and perpendicular to the inclined plane.

When $\theta$ = 45$^\circ$, the net force that brings the block down is

${F_d} = mg\sin \theta - F\cos \theta = {1 \over {\sqrt 2 }} - {1 \over {\sqrt 2 }} = 0$.

Thus, the block is stationary if $\theta = 45^\circ $. When $\theta > 45^\circ $, the force ${F_d} > 0$, and hence the block has a tendency to move down. Thus, the frictional force f acts on the block upwards i.e., towards Q.