The total spring constant of the system as shown in the figure will be

$A$ block of mass $M = 5 \ kg$ is suspended from one end of a spring. The spring extends by $l = 0.1 \ m$ due to the mass of the block. The block is given an upward velocity of $v = 2 \ m/s$. What is the maximum height $h$ (in $m$) reached by the block? $(g = 10 \ m/s^2)$

Initially,the system is in equilibrium. The time period of $SHM$ of the block in the vertical direction is

Define the torsional constant for a spring.

Find the maximum amplitude (in $cm$) of the $SHM$ such that block $A$ will not slip on block $B$. Given: spring constant $K = 100 \ N/m$,mass of block $A$ $m_A = 0.25 \ kg$,mass of block $B$ $m_B = 1.25 \ kg$,and coefficient of friction between $A$ and $B$ is $\mu = 0.4$. Take $g = 10 \ m/s^2$.

Two bodies of masses $1\, kg$ and $4\, kg$ are connected to a vertical spring,as shown in the figure. The smaller mass executes simple harmonic motion of angular frequency $25\, rad/s$,and amplitude $1.6\, cm$ while the bigger mass remains stationary on the ground. The maximum force exerted by the system on the floor is ..... $N$ (take $g = 10\, m/s^2$).