A block is kept on a frictionless inclined su | vedclass

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

Question

(B) Let the mass of the block be $m$. To keep the block stationary relative to the inclined surface,we analyze the forces in the frame of the incline.

Vedclass · Accepted Answer

$g 	an \alpha$

Vedclass · Answer

(B) Let the mass of the block be $m$. To keep the block stationary relative to the inclined surface,we analyze the forces in the frame of the incline.$1$. The forces acting on the block are its weight $(mg)$ acting downwards,the normal force $(N)$ perpendicular to the surface,and the pseudo force $(ma)$ acting horizontally in the direction opposite to the acceleration of the incline.$2$. Resolving the forces along the inclined surface:- The component of the gravitational force down the incline is $mg \sin \alpha$.- The component of the pseudo force up the incline is $ma \cos \alpha$.$3$. For the block to remain stationary relative to the incline,these two forces must balance each other:$ma \cos \alpha = mg \sin \alpha$$4$. Dividing both sides by $m \cos \alpha$:$a = g \frac{\sin \alpha}{\cos \alpha}$$a = g 	an \alpha$

$A$ block is kept on a frictionless inclined surface with an angle of inclination '$\alpha$'. The incline is given a horizontal acceleration '$a$' to keep the block stationary relative to the incline. Then '$a$' is equal to:

Similar Questions

$A$ vessel containing water is moving with a constant speed towards the right along a straight horizontal path. If the vessel is given a constant retardation towards the right along a straight line,which of the following diagrams represents the surface of the liquid?

If a body of mass $m$ is carried by a lift moving with an upward acceleration $a$,then the forces acting on the body are:
$(i)$ The reaction $R$ on the floor of the lift acting upwards.
$(ii)$ The weight $mg$ of the body acting vertically downwards.
The equation of motion will be given by:

$A$ spring balance is attached to the ceiling of a lift. $A$ man hangs his bag on the spring and the spring balance reads $49 \,N$, when the lift is stationary. If the lift moves downward with an acceleration of $5 \,m/s^2$, the reading of the spring balance will be $(g = 9.8 \,m/s^2)$. (in $\,N$)

$A$ man of mass $60 \ kg$ is standing in a lift moving up with a retardation of $2.8 \ ms^{-2}$. The apparent weight of the man is (in $N$)

$A$ coin is dropped in a lift. It takes time $t_1$ to reach the floor when the lift is stationary. It takes time $t_2$ when the lift is moving up with constant acceleration. Then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module