Initially,the spring is at its natural length and both blocks are at rest. Determine the maximum extension in the spring. Given $k = 20 \ N/m$,$m_1 = 0.5 \ kg$,$m_2 = 1 \ kg$,and $F = 1 \ N$.

$A$ spring of force constant $k$ is cut into two equal halves. The force constant of each half is

$A$ spring has length $l$ and force constant $K$. If it is cut into two springs of length $l_1$ and $l_2$ such that $l_1 = n l_2$ ($n$ is an integer). The force constant of the spring of length $l_2$ is

To the free end of a spring hanging from a rigid support,a block of mass $m$ is hung and slowly allowed to come to its equilibrium position. Then the stretching in the spring is $d$. If the same block is attached to the same spring and allowed to fall suddenly,the amount of maximum stretching is: (force constant,$k$)

Two plates each of mass $m$ are connected by a massless spring as shown below. $A$ weight $W$ is placed on the upper plate which compresses the spring further. When $W$ is removed,the entire assembly jumps up. The minimum weight $W$ needed for the assembly to jump up when the weight is removed is just more than ...........$m$.

$A$ spring of force constant $k$ is cut into lengths of ratio $1:2:3$. They are connected in series and the new force constant is $k'$. Then they are connected in parallel and the force constant is $k''$. Then $k':k''$ is