Question

An inclined plane making an angle of ${30}^{\circ }$ with the horizontal is placed in a uniform horizontal electric field $200\text{N/C}$ as shown in the figure. A body of mass $1\text{kg}$ and charge $5\text{mC}$ is allowed to slide down from rest from a height of $1\text{m}$. If the coefficient of friction is $0.2$, find the time taken by the body to reach the bottom.

$[g=9.8{\text{m/s}}^2;\sin {30}^{\circ }=\frac{1}{2};\cos {30}^{\circ }=\frac{\sqrt{3}}{2}]$

• A. $2.3\text{s}$
• B. $0.46\text{s}$
• C. $1.3\text{s}$
• D. $0.92\text{s}$

Answer 1

The correct option is C $1.3\text{s}$


Net force along the inclined plane.

$F=mg\sin \theta -qE\cos \theta -f$

$\Rightarrow F=mg\sin \theta -qE\cos \theta -\mu N$

$\Rightarrow F=mg\sin \theta -qE\cos \theta -\mu (qE\sin \theta +mg\cos \theta )$

$\Rightarrow F=1\times 9.8\times \frac{1}{2}-5\times {10}^{-3}\times 200\times \frac{\sqrt{3}}{2}-0.2(5\times {10}^{-3}\times 200\times \frac{1}{2}+1\times 9.8\times \frac{\sqrt{3}}{2})$

$\Rightarrow F\approx 2.24\text{N}$

So, $a\approx F/m\approx 2.24{\text{m/s}}^2$

Further,

$\frac{h}{\sin \theta }=0+\frac{1}{2}a{t}^2$

$\Rightarrow \frac{1}{1/2}=\frac{1}{2}\times 2.24\times t^2$

$\Rightarrow t\approx 1.3\text{s}$