$A$ system consists of two springs connected in series,each having a spring constant of $10 \text{ N m}^{-1}$. The minimum work required to stretch this system by $1 \text{ cm}$ in $\text{erg}$ is

$A$ body of mass $0.15 \ kg$ moving with a velocity of $15 \ ms^{-1}$ comes to rest when it hits a spring that is fixed at the other end. If the force constant of the spring is $1500 \ Nm^{-1}$,then the compression in the spring is: (in $m$)

$A$ light spring of length $10 \ cm$ is stretched by $2 \ cm$ when a mass of $20 \ g$ is attached to its end. The mass is pulled until the total length of the spring becomes $14 \ cm$. What is the elastic potential energy stored in the spring (in Joules)?

To simulate car accidents,auto manufacturers study the collisions of moving cars with mounted springs of different spring constants. Consider a typical simulation with a car of mass $1000 \; kg$ moving with a speed $18.0 \; km/h$ on a smooth road and colliding with a horizontally mounted spring of spring constant $6.25 \times 10^{3} \; N m^{-1}$. What is the maximum compression of the spring in $m$?

$K$ is the force constant of a spring. The work done in increasing its extension from $l_1$ to $l_2$ will be

When a $4\, kg$ mass is hung vertically on a light spring that obeys Hooke's law,the spring stretches by $2\, cm$. The work required to be done by an external agent in stretching this spring by $5\, cm$ will be ......... $J$ $(g = 9.8\, m/s^2)$.