$A$ man of mass $80\, kg$ is standing in an elevator which is moving with an acceleration of $6\, m/s^2$ in the upward direction. The apparent weight of the man will be ......... $N$ $(g = 10\, m/s^2)$.

$A$ wedge $Y$ with mass of $10 \text{ kg}$ has all frictionless surfaces,and the inclined surface makes an angle of $37^{\circ}$ with the horizontal. $A$ block $X$ with mass $2 \text{ kg}$ is placed at the highest point of the wedge as shown in the figure and is at rest. At $t=0$,the wedge $Y$ is pulled toward the right with a constant force $f$ of $24 \text{ N}$. Taking the block $X$ at rest at $t=0$,the time taken by it to slide down $8.8 \text{ m}$ on the slope,while $Y$ is in motion,is . . . . . . s. (Take $\tan(37^{\circ}) = 3/4$ and $g = 10 \text{ m/s}^2$)

$A$ man of weight $W$ is standing on a lift which is moving downward with an acceleration $a$. The apparent weight of the man is

$A$ vehicle is moving with a constant speed of $10 \ m/s$ on a circular horizontal track of radius $20 \ m$. $A$ bob is suspended from the roof of the vehicle by a massless string. The angle made by the string with the vertical will be (acceleration due to gravity,$g = 10 \ m/s^2$).

$A$ spring balance is attached to the ceiling of a lift. $A$ man hangs his bag on the spring and the spring 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 .......... $N$.

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: