> Q. An inclined plane making an angle of $30^{circ}$ with the horizontal is placed in a uniform horizontal electric field $200 frac{ N }{ C }$ as shown in the figure. A body of mass $1 kg$ and charge $5 mC$ is allowed to slide down from rest at a height of $1 m$. If the coefficient of friction is $0.2,$ find the time taken by the body to reach the bottom. $left[ g =9.8 m / s ^{2}, sin 30^{circ}=frac{1}{2}right.$; $left.cos 30^{circ}=frac{sqrt{3}}{2}right]$ – LIVE ANSWER TODAY

Q. An inclined plane making an angle of $30^{circ}$ with the horizontal is placed in a uniform horizontal electric field $200 frac{ N }{ C }$ as shown in the figure. A body of mass $1 kg$ and charge $5 mC$ is allowed to slide down from rest at a height of $1 m$. If the coefficient of friction is $0.2,$ find the time taken by the body to reach the bottom. $left[ g =9.8 m / s ^{2}, sin 30^{circ}=frac{1}{2}right.$; $left.cos 30^{circ}=frac{sqrt{3}}{2}right]$

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An inclined plane making an angle of $30^{circ}$ with the horizontal is placed in a uniform horizontal electric field $200 frac{ N }{ C }$ as shown in the figure. A body of mass $1 kg$ and charge $5 mC$ is allowed to slide down from rest at a height of $1 m$. If the coefficient of friction is $0.2,$ find the time taken by the body to reach the bottom. $left[ g =9.8 m / s ^{2}, sin 30^{circ}=frac{1}{2}right.$; $left.cos 30^{circ}=frac{sqrt{3}}{2}right]$

Solution:

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here $N =9.8 cos 30+1 sin 30$

$approx 9 N$

so $a =frac{9.8 sin 30-1 cos 30-mu N }{1}$

$a =2.233 m / s ^{2}$

By $S = ut +frac{1}{2} at ^{2}$

$=frac{1}{2}(2.233) t ^{2}$

$sin 30^{circ}$

$t approx 1.3 sec$

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