The work done in joules in increasing the extension of a spring of stiffness $10\, N/cm$ from $4\, cm$ to $6\, cm$ is:

A spring when stretched by $2 \,mm$ its potential energy becomes $4 \,J$. If it is stretched by $10 \,mm$, its potential energy is equal to

Two block of masses $m_1$ and $m_2$ connected with the help of a spring of spring constant $k$ initially to natural length as shown. A sharp impulse is given to mass $m_2$ so that it acquires a velocity $v_0$ towards right. If the system is kept an smooth floor then find the maximum elongation that the spring will suffer

Pulley and spring are massless and the friction is absent everwhere. $5\, kg$ block is  released from rest. The speed of $5\, kg$ block when $2\, kg$ block leaves the contact  with ground is (take force constant of the spring $K = 40\, N/m$ and $g = 10\, m/s^2)$

$A$ small block of mass $m$ is placed on $a$ wedge of mass $M$ as shown, which is initially at rest. All the surfaces are frictionless . The spring attached to the other end of wedge has force constant $k$. If $a'$ is the acceleration of $m$ relative to the wedge as it starts coming down and $A$ is the acceleration acquired by the wedge as the block starts coming down, then 

The block of mass $M$  moving on the frictionless horizontal surface collides with the spring of spring constant $K$ and compresses it by length $L$. The maximum momentum of the block after collision is