Question

An inclined plane whose length is $\sqrt{2}$ times $32.36\, m$ making an angle of $45^{\circ}$ with the horizontal is placed in an uniform electric field $E =150\, Vm ^{-1}$. A particle of mass $2\, kg$ 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

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)]
$\therefore 2 \times a=\frac{2 \times 10}{\sqrt{2}}+\frac{10^{-3} \times 150}{\sqrt{2}} -0.2\left(\frac{2 \times 10}{\sqrt{2}}-\frac{10^{-3} \times 150}{\sqrt{2}}\right)$
$\therefore 2 a =\frac{20}{\sqrt{2}}+\frac{150 \times 10^{-3}}{\sqrt{2}}-0.2\left(\frac{20}{\sqrt{2}}-\frac{150 \times 10^{-3}}{\sqrt{2}}\right)$
$\therefore 2 a =\frac{20.15}{\sqrt{2}}-0.2\left(\frac{19.85}{\sqrt{2}}\right)$
$\therefore 2 a =\frac{20.15}{\sqrt{2}}-\frac{3.97}{\sqrt{2}}$
$\therefore 2 a =\frac{16.18}{\sqrt{2}}$
$\therefore a =\frac{16.18}{2 \sqrt{2}} m / s ^{2}$
$s = ut +\frac{1}{2} at ^{2}$
as $u =0$,
$t =\sqrt{\frac{2 s }{ a }}=\sqrt{\frac{2 \times \sqrt{2} \times 32.36 \times 2 \sqrt{2}}{16.18}}=4\, s$