A body of mass 5 X 10^-3 kg is launched upon a rough inclined plane making an angle of 30° with the horizontal. Obtain the coefficient of friction between the body and the plane, if the time of ascent is half of the time of descent.    

According to question, the situation can be shown as

Case 1 For body projected up the plane,   $v=0$

Using third equation of motion,  $v^2=u^2-2a_{1}s$

$\Rightarrow 0=u^2-2a_{1}s$

$\therefore u=\sqrt{2a_{1}s}$

Also    $v=u-a_1{t}_1$

$\Rightarrow 0=u-a_1{t}_1$

$\Rightarrow t_1=u/a_1$

$\therefore t_1=\frac{\sqrt{2a_{1}s}}{a_1}=\sqrt{\frac{2s}{a_1}}$                                                   …(i)

$\text{ As, }m{a}_1=\mu R+mg\sin \theta$

$\therefore$   Retardation,  $a_1=\frac{\mu R+mg\sin 30°}{m}$

Also,  $R=mg\cos 30°$

$\Rightarrow a_1=\frac{\mu mg\cos 30°+mg\sin 30°}{m}$

$=\mu g\frac{\sqrt{3}}{2}+\frac{g}{2}$                                                             …(ii)

Case 2 For body coming down the plane,

$s=ut+\frac{1}{2}{a}_2{t}_2^2$

As     u = 0

$\Rightarrow t_2=\sqrt{2s/a_2}$                                                      …(iii)

Downward acceleration,

$a_2=\frac{mg\sin 30°-\mu R}{m}$

$=\frac{mg\sin 30°-\mu mg\cos 30°}{m}$

$=\frac{g}{2}-\frac{\mu g\sqrt{3}}{2}$                                                       …(iv)

Given,     $t_1=t_2/2$

Substituting values of t₁ and t₂ from Eqs. (i) and (iii), we get

$\sqrt{\frac{2s}{a_1}}=\frac{1}{2}\sqrt{\frac{2s}{a_2}}$

Now, squaring both sides, we get

$a_1=4a_2$                                                                …(v)

Substituting values of a₁ and a₂ from Eqs. (ii) and (iv) in Eq. (v), we get

$\mu g\frac{\sqrt{3}}{2}+\frac{g}{2}=4\left(\frac{g}{2}-\frac{\mu g\sqrt{3}}{2}\right)$

Solving $\mu$, we get

$\mu =\frac{\sqrt{3}}{5}=\frac{1.732}{5}$

$\Rightarrow \mu =0.346$