#### Answer

$P_i=483KW$

#### Work Step by Step

We can determine the required power supplied as follows:
$\Sigma F_x=ma_x$
$\implies F-F_D=ma$
We plug in the known values to obtain:
$F-(0.3)(28)^2=2300(5)$
$\implies F=11735.2N$
The efficiency of the motor is given as
$\epsilon=\frac{P_o}{P_i}$
We plug in the known values to obtain:
$0.68=\frac{328.58}{P_i}$
This simplifies to:
$P_i=483KW$