The spring balance inside a lift suspends an object. As the lift begins to ascend,the reading indicated by the spring balance will

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

$A$ block of mass $m$ is in contact with the cart $C$ as shown in the figure. The coefficient of static friction between the block and the cart is $\mu$. The acceleration $\alpha$ of the cart that will prevent the block from falling satisfies:

$A$ block of mass $m$ moving with a velocity $v_0$ on a smooth horizontal surface strikes and compresses a spring of stiffness $k$ until the mass comes to rest,as shown in the figure. This phenomenon is observed by two observers:
$A$: standing on the horizontal surface
$B$: standing on the block
To an observer $B$,when the block is compressing the spring:

$A$ man is standing in a lift which goes up and comes down with the same constant acceleration. If the ratio of the apparent weights in the two cases is $2 : 1$,then the acceleration of the lift is ......... $m/s^2$.

$A$ person standing on a spring balance inside a stationary lift measures $60 \, kg$. The weight of that person if the lift descends with a uniform downward acceleration of $1.8 \, m/s^{2}$ will be $_{-} N$. $[g = 10 \, m/s^{2}]$