NLM

$f_{max} = \mu mg \cos \theta = 0.8 \times 2 \times 10 \times = 13.6 N$
$\mathrm{F} = \mathrm{mg} \sin \theta = 2 \times 1 0 \times \frac {\sqrt {3}}{2} = 1 0 \mathrm{N}$
$\because \quad \mathrm{F} < \mathrm{f} _ {\max}, \mathrm{So} \mathrm{f} = \mathrm{F} = 1 0 \mathrm{N}$

$\mathrm{F_1} = \mathrm{mg}\sin \theta +\mu \mathrm{mg}\cos \theta$
$\mathrm{F} _ {2} = \mathrm{mg} \sin \theta - \mu \mathrm{mg} \cos \theta$
$\mathrm{F} _ {1} = 3 \mathrm{F} _ {2}$
$\therefore \quad \tan \theta + \mu = 3 (\tan \theta - \mu)$
$2 \mu = \tan \theta \Rightarrow \theta = \tan^ {- 1} (2 \mu)$

$a = g \sin \theta - \mu g \cos \theta$
$\begin{array}{r l} \mathrm{g} & \left(\frac {1}{\sqrt {2}} - 0. 2 5 \times \frac {1}{\sqrt {2}}\right) = \frac {1 0 \times 0 . 7 5}{\sqrt {2}} = \frac {7 . 5}{\sqrt {2}} \\ & = 5. 3 5 \mathrm{m} / \mathrm{s} ^ {2} \end{array}$

$f = \mu N$
$f = \mu mg \cos\theta$
$F_{\text{ext}} = mg \sin\theta + f$
$F_{\text{ext}} = mg \sin\theta + \mu mg \cos\theta$

mg sinθ = μ mgcosθ
$\sin \theta = \mu \cos \theta$
$\dots (\mathrm{i})$
when pushed up acceleration
$a = - (g \sin \theta + \mu g \cos \theta)$
$= - (\mathrm{g} \sin \theta + \mathrm{g} \sin \theta) = - 2 \mathrm{g} \sin \theta$
$\therefore \quad \mathrm{v} ^ {2} - \mathrm{v} _ {0} ^ {2} = 2 (\mathrm{a}) \times \mathrm{s}$
Final velocity v = 0
$- \mathrm{v} _ {0} ^ {2} = 2 \text {as} \Rightarrow - \mathrm{v} _ {0} ^ {2} = 2 (- 2 \mathrm{g} \sin \theta) \mathrm{s}$
$\therefore \quad s = \frac {v _ {0} ^ {2}}{2 (2 g \sin \theta)} = \frac {v _ {0} ^ {2}}{4 g \sin \theta}$

Let the length of incline is d
Case I : for rough incline plane
$a _ {r} = g \sin 4 5 ^ {\circ} - \mu g \cos 4 5 ^ {\circ} = \frac {g - \mu g}{\sqrt {2}} = \left(\frac {1 - \mu}{\sqrt {2}}\right) g$
time taken to slide down $(\mathrm{t_r}) = \sqrt{\frac{2\mathrm{d}}{\mathrm{a_r}}}$
Case II : For smooth incline plane
$a _ {s} = g \sin 4 5 ^ {\circ} = \frac {g}{\sqrt {2}}$
$\Rightarrow \quad t _ {s} = \sqrt {\frac {2 d}{a _ {s}}}$
According to question
$\mathsf {n t} _ {\mathrm{s}} = \mathsf {t} _ {\mathrm{r}} \quad \Rightarrow \quad \mathsf {n} ^ {2} \mathsf {t} _ {\mathrm{s}} ^ {2} = \mathsf {t} _ {\mathrm{r}} ^ {2}$
$\mathrm{n} ^ {2} \left(\frac {2 \mathrm{d}}{\mathrm{g} / \sqrt {2}}\right) = \frac {2 \mathrm{d}}{\left(\frac {1 - \mu}{\sqrt {2}}\right) \mathrm{g}}$
$\mathrm{n} ^ {2} = \frac {1}{1 - \mu} \Rightarrow \mu = 1 - \frac {1}{\mathrm{n} ^ {2}}$

Given

as angle made by normal reaction with vertical = angle of inclined plane = $\theta$ and block has to be pushed down,
So, $\theta <$ angle of repose.

Angle of repose is independent of mass.

$\tan \theta = \mu_{s}$
$\tan \theta = \sqrt {3}$
$\theta = 6 0 ^ {\circ}$

Given

$\therefore \quad \tan \theta = \mu_ {s}$
$3 / 4 = \mu_ {\mathrm{s}}$
$\mu_ {s} = 0. 7 5$

mg - Q cosθ = N
$\mathrm{P} + \mathrm{Q} \sin \theta = \mu \mathrm{N}$
$\therefore \quad \mu = \frac {\mathrm{P} + \mathrm{Q} \sin \theta}{\mathrm{mg} - \mathrm{Q} \cos \theta}$

$\mathrm {T^ {\prime} cos6 0 ^ {\circ} = W}$
$\mathrm {T^ {\prime} sin6 0 ^ {\circ} = T}$
$\therefore \tan 6 0 ^ {\circ} = \frac {T}{W} \Rightarrow W = \frac {T}{\sqrt {3}}$
$\mathrm{W} = \frac {3 4 6}{1 . 7} = 2 0 0 \mathrm{N}$

Given


$\mathrm {f_ {1max} = 0.5\times 80 = 40N}$
$\mathrm{f} _ {2 \max} = 0. 2 \times 1 0 0 = 2 0 \mathrm{N}$
∴ combined acceleration (a) = $\frac{60-20}{10} = 4 \, m/s^{2}$
from FBD of 8 kg :-
Friction between A & B
$\mathrm{f} _ {1} = 6 0 - 8 \times 4 = 2 8 \mathrm{N}$

Given
$0. 3 \times 4 \times 1 0 = 1 2 \xleftarrow [ \leftarrow a ]{\boxed {4 k g}}$
$\mathrm{F} \xleftarrow {\text { B }} 6 \mathrm{kg} \xleftarrow {\text { a }} 1 2 \mathrm{N}$
$\mathrm{a} = \frac {1 2}{4} = 3 \mathrm{m} / \mathrm{s} ^ {2}$
$\mathrm{F} - 1 2 = 6 \times \mathrm{a}$
$\mathrm{F} = 1 2 + 6 \times 3 = 3 0 \mathrm{N}$

Acceleration of system = $\frac{27}{9+3}=\frac{9}{4}$ m/s $^{2}$
For upper block w.r.t lower block
$\boxed {3 \mathrm{kg}} \rightarrow \frac {9}{4} \mathrm{m} / \mathrm{s} ^ {2}$
$\mathrm {f_ {\max}} = \frac {1}{3} \times 3 0 = 1 0 \mathrm{N}$
$\mathrm {f_ {acting}} = 3 \times \frac {9}{4} = 6. 7 5 \mathrm{N}$
$f _ {\mathrm{acting}} < f _ {\mathrm{max}}$
$\mathrm{So}, \mathrm{f} = \mathrm {f_ {acting}} = 6. 7 5 \mathrm{N}$

Acceleration
$= \frac{\text{Net force in the direction of motion}}{\text{Total mass of system}}$ $= \frac{\mathrm{m}_1\mathrm{g} - \mu(\mathrm{m}_2 + \mathrm{m}_3)\mathrm{g}}{\mathrm{m}_1 + \mathrm{m}_2 + \mathrm{m}_3}$ $= \frac{\mathrm{g}}{3}(1 - 2\mu)$

Change in momentum, change in momentum, $\Delta p = \int Fdt$ $= \text{Area of F-t graph}$ $= (\frac{1}{2} \times 2 \times 6) - (3 \times 2) + (4 \times 3)$ $= 12\mathrm{N-s}$

Let upthrust of air be $F_{a}$ then for downward motion
mg - $F_{a} = ma$ ...(1)
for upward motion (when $\Delta m$ is removed) $F_{a} - (m - \Delta m)g = (m - \Delta m)a$ ...(2)
After solving equation (1) & (2)
Therefore $\Delta m = \frac{2ma}{g + a}$

For the motion of both blocks
$\mathrm{m} _ {2} \mathrm{g} - \mathrm{T} = \mathrm{m} _ {2} \mathrm{a}$
$\mathrm{T} - \mu_ {\mathrm{k}} \mathrm{m} _ {1} \mathrm{g} = \mathrm{m} _ {1} \mathrm{a}$
$a = \frac {(m _ {2} - \mu_ {k} m _ {1}) g}{m _ {1} + m _ {2}}$
For the block of mass 'm $_{2}$ '
$\mathbf {m} _ {2} \mathbf {g} - \mathbf {T} = \mathbf {m} _ {2} \left[ \frac {\mathbf {m} _ {2} - \boldsymbol {\mu} _ {\mathrm{k}} \mathbf {m} _ {1}}{\mathbf {m} _ {1} + \mathbf {m} _ {2}} \right] \mathbf {g}$
$\mathbf {T} = \mathbf {m} _ {2} \mathbf {g} - \left[ \frac {\mathbf {m} _ {2} - \boldsymbol {\mu} _ {\mathrm{k}} \mathbf {m} _ {1}}{\mathbf {m} _ {1} + \mathbf {m} _ {2}} \right] \mathbf {m} _ {2} \mathbf {g}$
$\mathbf {T} = \mathbf {m} _ {2} \mathbf {g} \left[ \frac {\mathbf {m} _ {1} + \boldsymbol {\mu} _ {\mathrm{k}} \mathbf {m} _ {1}}{\mathbf {m} _ {1} + \mathbf {m} _ {2}} \right]$
$\mathrm{T} = \frac {\mathrm{m} _ {1} \mathrm{m} _ {2} (1 + \mu_ {\mathrm{k}}) \mathrm{g}}{\mathrm{m} _ {1} + \mathrm{m} _ {2}}$

Acceleration of system = $\frac{F_{net}}{M_{total}}$ $=\frac{14}{4+2+1}=2\ m/s^{2}$ 14 N → 4kg 2kg 1kg
The contact force between 4 kg & 2 kg block will move 2 kg & 1 kg block with the same acceleration
so $F_{contact} = (2 + 1)a = 3(2) = 6N.$

Coefficient of static friction,
$\mu_ {\mathrm{s}} = \tan 3 0 ^ {\circ} = \frac {1}{\sqrt {3}} = 0. 6$
$\mathrm{a} = \mathrm{g} \sin 3 0 ^ {\circ} - \mu_ {\mathrm{K}} \mathrm{g} \cos 3 0 ^ {\circ}$
$S = u t + \frac {1}{2} a t ^ {2}$
$4 = \frac {1}{2} \left[ \frac {\mathrm{g}}{2} - \frac {\mu_ {\mathrm{k}} \mathrm{g} \sqrt {3}}{2} \right] \times 1 6 \Rightarrow \mu_ {\mathrm{k}} = 0.5$

Impulse= $\left|\overrightarrow{\Delta p}\right|=m\left|\overrightarrow{\Delta V}\right|=m(2V\cos60^{\circ})=mV$

Before cutting the string :-

$\therefore \quad \mathrm{T} = \mathrm{mg}$
After cutting the string :-

$a _ {\mathrm{A}} = \frac {4 \mathrm{mg} - 3 \mathrm{mg}}{3 \mathrm{m}} = \frac {\mathrm{g}}{3}$
$a _ {B} = \frac {m g}{m} = g$

Coefficient of sliding friction has no dimension