(2)

Given: Angle of inclination, $\theta =45°$

$t_{\text{rough}}=2t_{\text{smooth}}$

Length of both inclined surfaces is equal.

For smooth surface, initial velocity is zero, i.e., $u=0,$ the distance $\text{'}S\text{'}$ is given by

       $S=\frac{1}{2}{a}_{\text{smooth}}{t}_{\text{smooth}}^2$

Acceleration of smooth plane is $a_{\text{smooth}}=g\sin \theta$

Using $\theta =45°$, we get

         $a_{\text{smooth}}=g\sin 45°=\frac{g}{\sqrt{2}}$ or $\frac{g\sqrt{2}}{2}$

$\Rightarrow S=\frac{1}{2}g\frac{\sqrt{2}}{2}{t}_{\text{smooth}}^2 \Rightarrow t_{\text{smooth}}=\sqrt{\frac{4S}{g\sqrt{2}}}$

For rough surface : 

$S=\frac{1}{2}{a}_{\text{rough}}{t}_{\text{rough}}^2 \text{(∵ }u=0\text{)}$

$a_{\text{rough}}=g\sin \theta -{\mu }_{k}g\cos \theta$

Using $\theta =45°$, we get $a_{\text{rough}}=\frac{g\sqrt{2}}{2}-{\mu }_{k}g\frac{\sqrt{2}}{2}$

$\therefore$  $S=\frac{1}{2}(g\frac{\sqrt{2}}{2}-{\mu }_{k}g\frac{\sqrt{2}}{2})t_{\text{rough}}^2$

or $t_{\text{rough}}=\sqrt{\frac{2S}{{\displaystyle g}\frac{\sqrt{2}}{2}(1-{\mu }_k)}}=\sqrt{\frac{4S}{g\sqrt{2}(1-{\mu }_k)}}$

As $t_{\text{rough}}=2t_{\text{smooth}}$

      $\sqrt{\frac{4S}{g\sqrt{2}(1-{\mu }_k)}}=2\sqrt{\frac{4S}{g\sqrt{2}}}$

Squaring on both sides, we get

$\Rightarrow \frac{4S}{g\sqrt{2}(1-{\mu }_k)}=4·\frac{4S}{g\sqrt{2}} \Rightarrow 4=\frac{1}{1-{\mu }_k}$

$\Rightarrow 4(1-{\mu }_k)=1 \text{or} 1=4-4{\mu }_k$

     ${\mu }_k=\frac{3}{4}=0.75$

(3)

Common acceleration of the system,

$a=\frac{10\text{N}}{2\text{kg}+3\text{kg}}=\frac{10}{5}{\text{ms}}^{-2}=2{\text{ms}}^{-2}$

Let R be the contact force between 2 kg and 3 kg blocks.
The free body diagram of 2 kg block is shown in the figure.

[IMAGE 24]

The equation of motion is,

$10-R=2a \text{or} R=10-2\times 2=6\text{N}$

So, the force exerted by block A on B = 6 N

(1)

Let $\mu$ be the coefficient of friction between the car and the road.

As, retarding force = force of friction

$\therefore$ $ma=\mu mg\Rightarrow a=\mu \times g$

Putting the given values, we get

$a=(0.15)\left(10{\text{ms}}^{-2}\right)=1.5{\text{ms}}^{-2}$

(4)

Coefficient of sliding friction has no dimension.

$f={\mu }_{s}N \Rightarrow {\mu }_s=\frac{f}{N}$

(4)

[IMAGE 26]------------------------------

Let ${\mu }_s$ and ${\mu }_k$ be the coefficients of static and kinetic friction between the box and the plank respectively.

When the angle of inclination $\theta$ reaches $30°$, the block just slides,

$\therefore {\mu }_s=\tan \theta =\tan 30°=\frac{1}{\sqrt{3}}=0.6$

If $a$ is the acceleration produced in the block, then

       $ma=mg\sin \theta -f_k$    (where $f_k$ is force of kinetic friction)

      $=mg\sin \theta -{\mu }_{k}N$                                      $\text{(as }{f}_k={\mu }_{k}N\text{)}$

      $=mg\sin \theta -{\mu }_{k}mg\cos \theta$                         $\text{(as }N=\mathrm{mgcos}\theta \text{)}$

       $a=g(\sin \theta -{\mu }_k\cos \theta )$

As    $g=10{\text{ms}}^{-2}$ and $\theta =30°$

$\therefore$$a=\left(10{\text{ms}}^{-2}\right)(\sin 30°-{\mu }_k\cos 30°)$                          ...(i)

If $s$ is the distance travelled by the block in time $t$, then

       $s=\frac{1}{2}a{t}^2 \text{(as }u=0\text{)} \text{or} a=\frac{2s}{t^2}$

But $s=4.0\text{m}$ and $t=4.0\text{s}$ (given)

$\therefore$ $a=\frac{2(4.0\text{m})}{(4.0{\text{s)}}^{{}^2}}=\frac{1}{2}{\text{ms}}^{-2}$

Substituting this value of $a$ in equation (i), we get

$\frac{1}{2}{\text{ms}}^{-2}=\left(10{\text{ms}}^{-2}\right)(\frac{{\displaystyle 1}}{{\displaystyle 2}}-{\mu }_k\frac{{\displaystyle \sqrt{3}}}{{\displaystyle 2}})$

$\frac{1}{10}=1-\sqrt{3}{\mu }_k \text{or} \sqrt{3}{\mu }_k=1-\frac{1}{10}=\frac{9}{10}=0.9$ or ${\mu }_k=\frac{0.9}{\sqrt{3}}=0.5$

(1)

(3)

[IMAGE 28]

Force of friction on mass $m_2=\mu m_{2}g$

Force of friction on mass $m_3=\mu m_{3}g$

Let $a$ be the common acceleration of the system.

$\therefore$  $a=\frac{m_{1}g-\mu m_{2}g-\mu m_{3}g}{m_1+m_2+m_3}$

Here,  $m_1=m_2=m_3=m$

$\therefore$  $a=\frac{mg-\mu mg-\mu mg}{m+m+m}$

$=\frac{mg-2\mu mg}{3m}$

$=\frac{g(1-2\mu )}{3}$