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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The forces acting on the block along the inclined plane are the component of gravity $mg \sin 30^{\circ}$ acting downwards,the frictional force $f

Vedclass · Accepted Answer

$2.3$

Vedclass · Answer

(B) The forces acting on the block along the inclined plane are the component of gravity $mg \sin 30^{\circ}$ acting downwards,the frictional force $f = \mu N$ acting upwards,and the component of the electric force $qE \cos 30^{\circ}$ acting upwards.First,we find the normal reaction $N$ on the block:$N = mg \cos 30^{\circ} + qE \sin 30^{\circ}$$N = (1 	imes 10 	imes \frac{\sqrt{3}}{2}) + (0.01 	imes 100 	imes \frac{1}{2}) = 5\sqrt{3} + 0.5 \approx 8.66 + 0.5 = 9.16 \ N$The net force $F_{net}$ along the incline is:$F_{net} = mg \sin 30^{\circ} - \mu N - qE \cos 30^{\circ}$$F_{net} = (1 	imes 10 	imes 0.5) - 0.2 	imes (9.16) - (0.01 	imes 100 	imes \frac{\sqrt{3}}{2})$$F_{net} = 5 - 1.832 - 0.866 = 2.302 \ N$Since $F_{net} = ma$,and $m = 1 \ kg$,the acceleration $a = 2.302 \ ms^{-2} \approx 2.3 \ ms^{-2}$.

Similar Questions

$A$ positively charged thin metal ring of radius $R$ is fixed in the $xy$-plane with its centre at the origin $O$. $A$ negatively charged particle $P$ is released from rest at the point $(0, 0, z_0)$,where $z_0 > 0$. Then the motion of $P$ is:

Electric charges $Q$ are placed on the $x$-axis at $x = 1, 2, 4, 8, \dots \text{meters}$ respectively. What are the electric field and electric potential at $x = 0$?

Three positive charges $q$ are placed at the vertices of an equilateral triangle. How do their electric field lines appear?

Vedclass Test Series

Exam Paper Generator

Online Exam Module