Question

An inclined plane making an angle ${30}^0$ with the horizontal is placed in a uniform horizontal electric field $200\raisebox{1ex}{$N$}\!\left/ \!\raisebox{-1ex}{$C$}\right.$ as shown in the figure. A body of mass of 1kg and the charge $5mC$ is allowed to slide down from rest from a height of $1m$. If the coefficient of friction is $0.2$, find the time taken by the body to reach the bottom.

• A. $2.3s$
• B. $0.46s$
• C. $1.3s$
• D. $0.92s$

Answer 1

The correct option is C

$1.3s$

Step 1: Given data.

Inclined angle, $\theta ={30}^0$

Uniform electric field, $E=200\raisebox{1ex}{$N$}\!\left/ \!\raisebox{-1ex}{$C$}\right.$

A body of mass, $m=1kg$

Charge, $q=5mC=5\times {10}^{-3}C$

Height, $h=1m$

Coefficient of friction, $\mu =0.2$

Step 2: Calculating net force

Free body diagram of the given figure:

Since the net force acting on the body down the plane.

$F_{net}=mg\sin \left(30\right)–(\mu R+qE\cos \left(30\right))$ ….$\left(i\right)$

Where $g$ is the acceleration due to gravity

From the free body diagram, the net force acting on the body is perpendicular to the plane.

$R=mg\cos \left(30\right)+qE\sin \left(30\right)$ ….$\left(ii\right)$

Substitute equation $\left(ii\right)$ in $\left(i\right)$ we get.

$F_{net}=mg\sin \left(30\right)–\mu \left(mg\cos \left(30\right)+qE\sin \left(30\right)\right)-qE\cos \left(30\right)$

$\Rightarrow$$F_{net}=mg\sin \left(30\right)–\mu mg\cos \left(30\right)-\mu qE\sin \left(30\right)-qE\cos \left(30\right)$

$\Rightarrow$$F_{net}=mg(\sin \left(30\right)–\mu \cos \left(30\right))-qE(\mu \sin \left(30\right)+\cos \left(30\right))$

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

$\Rightarrow$$F_{net}=2.3N$ ….$\left(iii\right)$

Step 3: Calculating distance

Again,

From the figure Distance traveled along the plane is:

$\sin \left(30\right)=\frac{h}{s}$

$\Rightarrow$ $s=2m$ …$\left(iv\right)$

Step 4: Calculating time

From the equation of motion:

$s=ut+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.a{t}^2$

Where $s$ is the displacement or distance traveled, $u$ is the initial velocity, $a$ is the acceleration, and $t$ is the time.

$\Rightarrow$$2=0\times t+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\left(\frac{F}{m}\right)t^2$ $\left[\because u\sin gF=ma\right]$

$\Rightarrow$$2=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\left(\frac{2.3}{1}\right)t^2$

$\Rightarrow$$t=1.3sec$

Hence, option C is correct.