Answer
(a) $2.9\times 10^4Kg$
(b) $3.8\times 10^7W$
Work Step by Step
(a) We know that
$W=\Delta K.E$
$\implies W=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2$
$\implies W=\frac{1}{2}mv_f^2-\frac{1}{2}m(0)^2=\frac{1}{2}mv_f^2$
This can be rearranged as:
$m=\frac{2W}{v_f^2}$
We plug in the known values to obtain:
$m=\frac{2(7.6\times 10^7)}{(72)^2}$
$m=2.9\times 10^4Kg$
(b) The required power can be calculated as
$P=\frac{W}{t}$
We plug in the known values to obtain:
$P=\frac{7.6\times 10^7}{2.0}$
$P=3.8\times 10^7W$
