Electrostatics Part 2

Given, mass of block, $m=1\,\text{kg}$ Acceleration due to gravity, $g=9.8\,\text{ms}^{-2}$ Inclination, $\theta=30^{\circ}$ Electric field, $E=200\,\text{N/C}$ Coefficient of friction, $\mu=0.2$ Charge, $q=5\,\text{mC}=5 \times 10^{-3}\,\text{C}$ Let friction force, $f=\mu N$ where, $N$ be the normal reaction. Since, net force is zero along the perpendicular direction of incline. Therefore, force along $Y$-axis will be zero. $\Rightarrow\;\; N = m g \cos 30^{\circ}+q E \sin 30^{\circ}$ $\Rightarrow\;\; N = 1 \times 9.8 \times \frac{\sqrt{3}}{2}+5 \times 10^{-3} \times 200 \times \frac{1}{2}$ $= 8.49+0.5$ $= 8.99\,\text{N} \approx 9\,\text{N}$ $\therefore\;\; f = \mu N = \frac{2}{10} \times 9 = \frac{18}{10} = 1.8\,\text{N}$ Now, total force along the plane of incline, $m g \sin 30^{\circ} - f - q E \cos 30^{\circ} = m a$ $\Rightarrow\;\; 1 \times 9.8 \times \frac{1}{2} - 1.8 - 5 \times 10^{-3} \times 200 \times \frac{\sqrt{3}}{2} = a$ $\Rightarrow\;\; 5 - 1.8 - \frac{1.732}{2} = a$ $\Rightarrow\;\; a = 2.34\,\text{ms}^{-2}$ Since, initial velocity of body, $u=0\,\text{ms}^{-1}$ and distance along incline, $s = \frac{h}{\sin 30^{\circ}} = \frac{1}{\sin 30^{\circ}} = 2$ By using second equation of motion, $s = u t + \frac{1}{2} a t^{2}$ $\Rightarrow\;\; 2 = 0 + \frac{1}{2} \times 2.34 \times t^{2}$ $\Rightarrow\;\; t^{2} = \frac{4}{2.34}$ $\Rightarrow\;\; t = \frac{2}{\sqrt{2.34}}$