Question

An inclined plane making an angle ${30}^o$ with the horizontal is placed in a uniform horizontal electric field of $100V{m}^{-1}$. A particle of mass $1kg$ and charge $0.01C$ is allowed to slide down from rest from the top of the inclined plane. If the coefficient of friction is $0.2$, the particle reaches the bottom of the inclined plane in $1$ second. Then the length of the inclined plane is (Acceleration due to gravity $=10{ms}^{-2}$)

• A. $1.150m$
• B. $1.323m$
• C. $1.151m$
• D. $1.172m$

Answer 1

The correct option is A $1.150m$

Direction of electric field is not given exactly but it is given that it is in horizontal plane. Two scenarios are possible as shown in the attached diagram.

Scenario (A)

Net force perpendicular to the incline is zero as the block has no displacement perpendicular to the inclined plane.

From the free body diagram,

$N+qE\sin \theta =mg\cos \theta$

$N=1\times 10\times \cos \left({30}^o\right)-0.01\times 100\times \sin \left({30}^o\right)$

$N=5\sqrt{3}-0.5N$

Frictional force (sliding) acting is:

$F_r=\mu N$

$F_r=0.2\times (5\sqrt{3}-0.5)$

$F_r=\sqrt{3}-0.01N$

From the free body diagram, net force down the incline is:

$F_d=mg\sin \theta +qE\cos \theta -F_r$

$F_d=1\times 10\times \sin \left({30}^o\right)+0.01\times 100\times \cos \left({30}^o\right)-(\sqrt{3}-0.01)$

$F_d=5.01-\frac{\sqrt{3}}{2}\approx 4.134N$

From newton's second law of motion, acceleration is:

$a=\frac{F_d}{m}=\frac{4.134}{1}=4.134{m/s}^2$

Using second equation of motion,

$s=ut+\frac{1}{2}a{t}^2$

$s=0\times 1+\frac{1}{2}\times 4.134\times 1^2$

$s=2.067m$

Scenario (B)

Net force perpendicular to the incline is zero as the block has no displacement perpendicular to the inclined plane.

From the free body diagram,

$N=mg\cos \theta +qE\sin \theta$

$N=1\times 10\times \cos \left({30}^o\right)+0.01\times 100\times \sin \left({30}^o\right)$

$N=5\sqrt{3}+0.5N$

Frictional force (sliding) acting is:

$F_r=\mu N$

$F_r=0.2\times (5\sqrt{3}+0.5)$

$F_r=\sqrt{3}+0.01N$

From the free body diagram, net force down the incline is:

$F_d=mg\sin \theta -qE\cos \theta -F_r$

$F_d=1\times 10\times \sin \left({30}^o\right)-0.01\times 100\times \cos \left({30}^o\right)-(\sqrt{3}+0.01)$

$F_d=4.99-\frac{3\sqrt{3}}{2}\approx 2.4N$

From newton's second law of motion, acceleration is:

$a=\frac{F_d}{m}=\frac{2.4}{1}=2.4{m/s}^2$

Using second equation of motion,

$s=ut+\frac{1}{2}a{t}^2$

$s=0\times 1+\frac{1}{2}\times 2.4\times 1^2$

$s=1.2m$