An inclined plane making an angle \(30^{\circ}\) with the horizontal is placed in a uniform horizontal…
An inclined plane making an angle \(30^{\circ}\) with the horizontal is placed in a uniform horizontal electric field of \(100 \mathrm{Vm}^{-1}\) as shown in the figure. A small block of mass \(\mathrm{I} \mathrm{kg}\) and charge, 0.01C is allowed to slide down from rest from a height, \(h=1 \mathrm{~m}\). If the coefficient of friction is 0.2 , then the acceleration of the block is nearly,
(Acceleration due to gravity, \(g=10 \mathrm{~ms}^{-2}\) )
\(1.3 \mathrm{~ms}^{-2}\)
\(2.3 \mathrm{~ms}^{-2}\)
\(3.3 \mathrm{~ms}^{-2}\)
\(4.3 \mathrm{~ms}^{-2}\)
Solution
According to the question, an inclined plane is making an angle of \(30^{\circ}\) with the horizontal, placed in a uniform electric field of \(100 \mathrm{~nm}^{-1}\). It can be such in the figure that a block of mass \(m\) is sliding down from rest at height \(h\).
From the above diagram, the total force \(F\) acting along inclined plane.
From fig,
\(m g \sin 30^{\circ}-\mu m g \cos 30^{\circ}-q E \cos 30^{\circ}=m a=F\)
Given,
\(\mu=0.2, m=1 \mathrm{~kg},, q=0.01 \mathrm{C}\) and \(h=1 \mathrm{~m}\)
Putting these values, we get
\(\begin{aligned}
& 10 \times \frac{1}{2}-0.2 \times 10 \times \frac{\sqrt{3}}{2}-0.01 \times 100 \times \frac{\sqrt{3}}{2}=a \\
& a=5-\sqrt{3}-0.5 \sqrt{3} \approx 2.3 \mathrm{~ms}^{-2}
\end{aligned}\)
Hence, the acceleration of the block is nearly, \(2.3 \mathrm{~ms}^{-2}\).
