Question

An inclined plane whose length is $\sqrt{2}$ times $32.36m$ making an angle of $45^{\circ }$ with the horizontal is placed in an uniform electric field $E=150V{m}^{-1}$. A particle of mass $2kg$ and charge $10^{-3}C$ is allowed to slide down from rest position from maximum height of slope. If the coefficient of friction is $0.2$, the time taken by the particle to reach the bottom is_________$s$

Answer 1

Answer: 4

As there is no vertical motion,
$N+qE\sin 45^{\circ }-mg\cos 45^{\circ }=0$
$\Rightarrow N+qE\sin 45^{\circ }=mg\cos 45^{\circ }$
$\Rightarrow N=\frac{mg}{\sqrt{2}}-\frac{qE}{\sqrt{2}}$ ...(i)
Considering horizontal motion with acceleration $a$,
$mg\sin 45^{\circ }+qE\cos 45^{\circ }-\mu N=ma$
$\therefore ma=\frac{mg}{\sqrt{2}}+\frac{qE}{\sqrt{2}}-\mu \left(\frac{mg}{\sqrt{2}}-\frac{qE}{\sqrt{2}}\right)$ ....[From (i)]
$\therefore 2\times a=\frac{2\times 10}{\sqrt{2}}+\frac{10^{-3}\times 150}{\sqrt{2}}-0.2\left(\frac{2\times 10}{\sqrt{2}}-\frac{10^{-3}\times 150}{\sqrt{2}}\right)$
$\therefore 2a=\frac{20}{\sqrt{2}}+\frac{150\times 10^{-3}}{\sqrt{2}}-0.2\left(\frac{20}{\sqrt{2}}-\frac{150\times 10^{-3}}{\sqrt{2}}\right)$
$\therefore 2a=\frac{20.15}{\sqrt{2}}-0.2\left(\frac{19.85}{\sqrt{2}}\right)$
$\therefore 2a=\frac{20.15}{\sqrt{2}}-\frac{3.97}{\sqrt{2}}$
$\therefore 2a=\frac{16.18}{\sqrt{2}}$
$\therefore a=\frac{16.18}{2\sqrt{2}}m/s^2$
$s=ut+\frac{1}{2}a{t}^2$
as $u=0$,
$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2\times \sqrt{2}\times 32.36\times 2\sqrt{2}}{16.18}}=4s$