Answer
$P_{operate}=68.125kW$
$P_{accelerate}=43.74kW$
Work Step by Step
First we calculate the total mass of the sistem:
$n_{chairs}=\frac{1000m}{20m}=50 chairs$
$m=50*250kg=12500kg$
The time that take to send all the chairs is:
$t=\frac{d}{v}=\frac{1km}{10\frac{km}{h}}*(\frac{3600s}{1h})=360s$
As the lift is operating at steady speed, the work done to raise the chairs is:
$W=mg\Delta z=12500kg*9.81\frac{m}{s^2}*200m=24525kJ$
Finally the power needed is:
$P=\frac{W}{t}=\frac{24525kJ}{360s}=68.125kW$
On the other hand, we need to calculate the power required to accelerate this ski lift in $5s$, this power will be sum of the power of acceleration and the power to raise the chairs.
The power of acceleration from rest to its operating speed will be:
$W_{a}=\frac{m}{2}\Delta V^2=(\frac{12500kg}{2})*((10\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s})^2-(0\frac{m}{s})^2)=48.225kJ$
$P_{a}=\frac{W}{t}=\frac{48.225kJ}{5s}=9.645kW$
The acceleration and vertical traveled during the acceleration will be:
$a=\frac{v}{t}=\frac{10\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}}{5s}=0.556\frac{m}{s^2}$
$h=(h_{0}+\frac{1}{2}at^2)sin(\theta)=(0m+\frac{1}{2}*0.556\frac{m}{s^2}*(5s)^2)*(\frac{200m}{1000m})=1.39m$
The power to raise the chairs that vertical distance will be:
$W_{p}=mg\Delta z=12500kg*9.81\frac{m}{s^2}*1.39m=170.45kJ$
$P_{p}=\frac{W}{t}=\frac{170.45kJ}{5}=34.09kW$
Finally the total power will be:
$P_{t}=P_{a}+P_{p}=9.645kW+34.09kW=43.74kW$
