Answer
$-8.4\times 10^6N$
Work Step by Step
We can find the required force as
$W=\Delta K.E$
$\implies Fd=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2$
$\implies Fd=\frac{1}{2}m(0)^2-\frac{1}{2}mv_i^2$
This can be rearranged as:
$F=\frac{-mv_i^2}{2d}$
We plug in the known values to obtain:
$F=-\frac{(27lb\times 1Kg/2.2lb)(550)^2}{2(0.22)}$
$F=-8.4\times 10^6N$
