In This post you will get answer of 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. $left[ g =9.8 m / s ^{2}, sin 30^{circ}=frac{1}{2}right.$; $left.cos 30^{circ}=frac{sqrt{3}}{2}right]$
Solution:
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 $S = ut +frac{1}{2} at ^{2}$
$=frac{1}{2}(2.233) t ^{2}$
$sin 30^{circ}$
$t approx 1.3 sec$
Solution:
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 $S = ut +frac{1}{2} at ^{2}$
$=frac{1}{2}(2.233) t ^{2}$
$sin 30^{circ}$
$t approx 1.3 sec$