Question

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

Answer 1

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