Question

An inclined plane makes an angle of $30^{\circ }$ with the horizontal electric field $E$ of $100V/m$. A particle of mass $1kg$ and charge $0.01C$ slides down from a height of $1m$. If the coefficient of friction is $0.2$, find the time taken for the particle to reach the bottom. (in sec)

Answer 1

Answer: 1.34

$f=\mu (qE\sin 30+mg\cos 30)$
$f=0.2\left(0.01\times 100\times \frac{1}{2}+1\times 10\times 0.866\right)$
$f=1.832N$
From above figure
$\frac{h}{S}=\sin 30^{\circ }$
$S=\frac{h}{\sin 30}=\frac{1}{1/2}$
$S=2m$
Net force along the inclined plane
$F=mg\sin 30-qE\cos 30-f$
$=1\times 10\times \frac{1}{2}-0.01\times 100\times 0.866-1.832)$
$=5-0.866-1.832=2.302N$
$a=\frac{F}{m}$
$=\frac{2.30}{1}=2.30m/s^2$
$S=ut+\frac{1}{2}a{t}^2(u=0)$
$S=\frac{1}{2}a{t}^2$
$t=\sqrt{\frac{2S}{a}}=\sqrt{\frac{2\times 2}{2.30}}$
$=1.32\sec$
$t=1.32\sec$