$A$ 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:

The ratio of the weight of a man in a stationary lift and when it is moving downward with uniform acceleration $a$ is $3 : 2$. The value of $a$ is ($g$ = acceleration due to gravity of the earth).

An elevator is moving vertically up with an acceleration $a$. The force exerted on the floor by a passenger of mass $m$ is:

$A$ man weighing $60 \text{ kg}$ is in a lift moving down with an acceleration of $1.8 \text{ ms}^{-2}$. The force exerted by the floor on him is:

In the figure shown,$P$ is a plate on which a wedge $B$ is placed,and on $B$ a block $A$ of mass $m$ is placed. The plate is suddenly removed and the system of $B$ and $A$ is allowed to fall under gravity. Neglecting any force due to air on $A$ and $B$,the normal force on $A$ due to $B$ is

Find the apparent weight of a body of mass $1.0 \,kg$ falling with an acceleration of $10 \,ms^{-2}$. $\left(g \approx 10 \,ms^{-2}\right)$