An inclined plane making an angle of $30^{0}$ with the horizontal is placed in a uniform horizontal electric field $200 \frac{N}{C}$ as shown in the figure. A body of mass 1kg and charge 5mc is allowed to slide down from rest at a height of 1m. If the coefficient of friction is 0.2, find the time taken by the body to reach the bottom $[g = 9.8 m / s^{2}, \sin 30^{0} = \frac{1}{2}, \cos 30^{0} = \frac{\sqrt{3}}{2}]$

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Detailed Solution

The force on the charge due to electric field, $F e = q E = 5 \times 10^{- 3} \times 200 = 1 N$
Normal reaction N = $m g \cos 30^{0} + q E \sin 30^{0}$
$= 1 \times 9.8 \times \frac{\sqrt{3}}{2} + 1 \times \frac{1}{2} ≃ 9 N$
Acceleration of the particle $a = \frac{m g \sin 30^{0} - q E \cos 30^{0} - μ N}{m}$
$= \frac{1 \times 9.8 \times \frac{1}{2} - 1 \times \frac{\sqrt{3}}{2} - 0.2 \times 9}{1} = 2.33 m / s^{2}$