# A spring with a spring constant of 250 N/m has a length of 0.5 meters at rest. What magnitude of force is needed to stretch the spring so that is 0.75 meters long?

Per Hooke's Law, in an ideal spring, the spring force F$_s$ is proportional to x where x is the displacement of the block from the equilibrium position
$F_s = -kx$.
$x =$ displacement of the block, i.,e, $x =0.75m - 0.5m = 0.25m$; It is given that the spring constant is 250Nm$^{-1}$
$\Rightarrow F_s = -k x = 250 Nm^{-1} \times 0.25\;m = 62.5 N$