An inclined plane making an angle of 30° with the horizontal is placed in a uniform horizontal electric field 200 $$\frac{N}{C}$$ 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.
[g = 9.8 m $$s^{-2}$$; $$\sin 30^{\circ} =\frac{1}{2}$$; $$\cos 30^{\circ} = \frac{\sqrt{3}}{2}$$] 

$$s = \frac{h}{\sin 30^\circ} = \frac{1}{1/2} = 2\ \text{m}$$

$$N = mg \cos 30^\circ + qE \sin 30^\circ$$

$$F_{\text{net}} = mg \sin 30^\circ - qE \cos 30^\circ - \mu N$$

$$\implies F_{\text{net}} = mg \sin 30^\circ - qE \cos 30^\circ - \mu (mg \cos 30^\circ + qE \sin 30^\circ)$$

$$\implies a = \frac{F_{\text{net}}}{m} = g(\sin 30^\circ - \mu \cos 30^\circ) - \frac{qE}{m}(\cos 30^\circ + \mu \sin 30^\circ)$$

$$\implies a = 9.8\left(\frac{1}{2} - 0.2 \cdot \frac{\sqrt{3}}{2}\right) - \frac{5 \times 10^{-3} \times 200}{1}\left(\frac{\sqrt{3}}{2} + 0.2 \cdot \frac{1}{2}\right)$$

$$\implies a = 9.8(0.5 - 0.1\sqrt{3}) - 1(0.5\sqrt{3} + 0.1)$$ $$\implies a = 4.9 - 0.98\sqrt{3} - 0.5\sqrt{3} - 0.1 = 4.8 - 1.48\sqrt{3}$$ 

$$\implies a = 4.8 - 1.48(1.732) = 4.8 - 2.563 = 2.237\ \text{m/s}^2$$

$$t = \sqrt{\frac{2s}{a}} = \sqrt{\frac{2(2)}{2.237}} = \sqrt{\frac{4}{2.237}} \approx 1.34\ \text{s}$$