Answer
Minimum power needed: $P_{p}=15.60kW$
Velocity doubled: $P_{p}=31.20kW$
Work Step by Step
In this case we are working with potential energy, then the potential work will be:
$W_{p}=mgh$
And the power:
$P_{p}=\frac{W_{p}}{t}=\frac{mgh}{t}=mgV_{vertical}$
The vertical velocity is:
$V_{vertical}=Vsin45^{\circ}=0.6\frac{m}{s}*sin45=0.424\frac{m}{s}$
And the total mass is:
$m=50*75kg=3750kg$
Then the minimum power input needed to drive this escalator is:
$P_{p}=3750kg*9.81\frac{m}{s^2}*0.424\frac{m}{s}=15.60kW$
If you want to doubled the velocity:
$P_{p}=3750kg*9.81\frac{m}{s^2}*(2*0.424\frac{m}{s})=31.20kW$
