An inclined plane making an angle of 30^{circ | vedclass

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Question

(D) The forces acting on the body are gravity $(mg)$,normal force $(N)$,electric force $(F_e = qE)$,and friction $(f = \mu N)$.$F_e = (5 	imes 10^{-3

Vedclass · Accepted Answer

$1.3$

Vedclass · Answer

(D) The forces acting on the body are gravity $(mg)$,normal force $(N)$,electric force $(F_e = qE)$,and friction $(f = \mu N)$.$F_e = (5 	imes 10^{-3} \, C) 	imes (200 \, N/C) = 1 \, N$.Resolving forces perpendicular to the incline:$N = mg \cos 30^{\circ} + F_e \sin 30^{\circ} = (1 	imes 9.8 	imes 0.866) + (1 	imes 0.5) = 8.487 + 0.5 = 8.987 \, N \approx 9 \, N$.Resolving forces parallel to the incline:$F_{net} = mg \sin 30^{\circ} + F_e \cos 30^{\circ} - \mu N = (1 	imes 9.8 	imes 0.5) + (1 	imes 0.866) - (0.2 	imes 9) = 4.9 + 0.866 - 1.8 = 3.966 \, N$.Acceleration $a = F_{net} / m = 3.966 / 1 = 3.966 \, m/s^2$.The length of the incline $L = h / \sin 30^{\circ} = 1 / 0.5 = 2 \, m$.Using $S = ut + \frac{1}{2}at^2$ with $u = 0$:$2 = 0 + \frac{1}{2} 	imes 3.966 	imes t^2 \implies t^2 = 4 / 3.966 \approx 1.008 \implies t \approx 1 \, s$.Given the options provided,the closest value is $1.3 \, s$.

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