Answer
(a) $0.0108m$
(b) $14.58J$
Work Step by Step
(a) We can find the required compression as follows:
$K=\frac{F}{x}$
$\implies x=\frac{F}{K}$
We plug in the known values to obtain:
$x=\frac{2700N}{2.5\times 10^5N/m}$
$x=0.0108m$
(b) We can find the required energy stored as follows:
$U=\frac{1}{2}Kx^2$
We plug in the known values to obtain:
$U=\frac{1}{2}(2.5\times 10^5N/m)(0.0108m)^2$
$U=14.58J$
