Answer
$P_i=622KW$
Work Step by Step
We can determine the required power as follows:
$\Sigma F_x=ma_x$
$F-(10v)=2300(6)$
$\implies F=13800+10v~~~$eq(1)
We know that
$v=v_{\circ}+at$
$v=0+(6)(5)=30m/s$
From eq(1),
$F=13800+10(30)=14100N$
The power output is given as
$P_o=Fv=14100\times 30=423KW$
The efficiency of the motor is
$\epsilon=\frac{P_o}{P_i}$
$\implies 0.68=\frac{423}{P_i}$
$\implies P_i=622KW$
