The weight of a man in a stationary lift is $w_1$ and when it is moving downwards with uniform acceleration $a$ is $w_2$. If the ratio $w_1 : w_2 = 4 : 3$,then the value of $a$ is ($g =$ acceleration due to gravity).

$A$ 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 $g = 10 \text{ m/s}^2$)

In the figure,the reading of the spring balance will be $......\,N$ $\left[ g = 10\,m/s^2 \right]$.

$A$ body of mass $2 \, kg$ is hung on a spring balance mounted vertically in a lift. If the lift descends with an acceleration equal to the acceleration due to gravity $g$,and if the lift moves up with a constant velocity of $2 \, m/s$,the reading on the balance will be ......... $kg$.

The ratio of the weight of a man in a stationary lift to his weight in a lift moving downwards with an acceleration '$a$' is $3:2$. Then '$a$' = . . . . . . .

Two wooden blocks of masses $M$ and $m$ are placed on a smooth horizontal surface as shown in the figure. If a force $P$ is applied to the system as shown in the figure such that the mass $m$ remains stationary with respect to the block of mass $M$,then the magnitude of the force $P$ is