Answer
a) $K=0.63N/m$
b) $U=7.2\times 10^{-6}J$
c) $F=0.0042N$
Work Step by Step
(a) We can find the required force constant as
$F=-Kx$
This can be rearranged as:
$K=\frac{-F}{x}$
We plug in the known values to obtain:
$K=\frac{-0.0030}{-0.0048}$
$K=0.63N/m$
(b) We can find the required energy stored as
$U=\frac{1}{2}Kx^2$
We plug in the known values to obtain:
$U=\frac{1}{2}(0.625)(0.0048)^2$
$U=7.2\times 10^{-6}J$
(c) We can find the required force as
$F=-Kx$
$F=-K(-\sqrt{\frac{2U}{K}})$ where $x=-\sqrt{\frac{2U}{K}}$
$F=\sqrt{2UK}$
We plug in the known values to obtain:
$F=\sqrt{2(2\times 7.2\times 10^{-6})(0.625)}$
$F=0.0042N$
