An inclined plane making an angle of $30^\circ$ with the horizontal is placed in a uniform horizontal electric field $200\ \frac{\text{N}}{\text{C}}$ as shown in the figure. A body of mass $1$ kg and charge $5$ mC is allowed to slide down from rest at a height of $1$ m. If the coefficient of friction is $0.2$, find the time taken by the body to reach the bottom. $\left[g = 9.8\ \text{m s}^{-2};\ \sin30^\circ = \frac{1}{2};\ \cos30^\circ = \frac{\sqrt{3}}{2}\right]$
Answer: (B) $1.3$ s
Electric force $qE = 5\times10^{-3}\times200 = 1$ N, horizontal, towards the rising side of the incline.
Components of $qE$: along the incline (up the slope) $qE\cos30^\circ = 0.866$ N; into the incline $qE\sin30^\circ = 0.5$ N.
Normal reaction: $N = mg\cos30^\circ + qE\sin30^\circ = 8.49 + 0.5 = 8.99$ N, so friction $= 0.2\times8.99 = 1.80$ N.
Net force down the slope: $mg\sin30^\circ - qE\cos30^\circ - f = 4.9 - 0.866 - 1.80 = 2.23$ N, so $a = 2.23$ m s$^{-2}$.
Length of incline $= \dfrac{1}{\sin30^\circ} = 2$ m.
$$t = \sqrt{\frac{2s}{a}} = \sqrt{\frac{4}{2.23}} \approx 1.3\ \text{s}$$
Solution by Sreeraj P, M.Sc Physics