Answer
$P_i=22.2KW$
Work Step by Step
We can determine the power supplied as follows:
$\Sigma F_y=0$
$\implies T_C+3T-m_Eg=0$
$\implies m_Cg+3T=m_Eg$
$\implies (60)(9.81)+3T=400(9.81)$
$\implies T=1111.8N$
We know that
$2v_E+(v_E-v_p)=0$
$\implies v_p=3v_E=3(4)=12m/s$
The power output is
$P_o=Fv=1111.8\times 12=13.34KW$
The efficiency of the motor is
$\epsilon=\frac{P_o}{P_i}$
We plug in the known values to obtain:
$0.6=\frac{13.34KW}{P_i}$
This simplifies to:
$P_i=22.2KW$
