As shown in figure there is a spring block system. Block of mass $500\,g$ is pressed against a horizontal spring fixed at one end to compress the spring through $5.0\,cm$ . The spring constant is $500\,N/m$ . When released, calculate the distance where it will hit the ground $4\,m$ below the spring ? $(g = 10\,m/s^2)$

A block is fastened to a horizontal spring. The block is pulled to a distance $x =10\,cm$ from its equilibrium position (at $x =0$ ) on a frictionless surface from rest. The energy of the block at $x =5$ $cm$ is $0.25\,J$. The spring constant of the spring is $.........Nm ^{-1}$

$A$ block of mass $m$ moving with a velocity $v_0$ on a smooth horizontal surface strikes and compresses a spring of stiffness $k$ till 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 

According to observer $B$, the potential energy of the spring increases

As shown in figure, two blocks are connected with a light spring. When spring was at its natural length, velocities are given to them as shown in figure. Choose the wrong alternative.

Two masses $A$ and $B$ of mass $M$ and $2M$ respectively are connected by a compressed ideal spring. The system is placed on $a$ horizontal frictionless table and given $a$ velocity $u\, \hat k$ in the $z$ -direction as shown in the figure. The spring is then released. In the subsequent motion the line from $B$ to $A$ always points along the $\hat i$ unit vector. At some instant of $\rho$ time mass $B$ has $a$ $x$ -component of velocity as $V_x\, \hat i$ . The velocity ${\vec V_A}$ of as $A$ at that instant is

The energy stored in wound watch spring is