Answer
a)$P_{p,full-counterweight}=4.71kW$
$P_{p,full-nocounterweight}=9.42kW$
b)$P_{p,empty}=2.94kW$
$P_{p,friction-empty}=3.90kW$
Work Step by Step
Here we are working with potential energy:
The potential work is: $W_{p}=mgh$
And the potential power is: $P_{p}=\frac{W_{p}}{t}=\frac{mgh}{t}=mgV$
The potential power at full load is:
$P_{p,full-counterweight}=(800-400)kg*9.81\frac{m}{s^2}*1.2\frac{m}{s}=4.71kW$
The potential power with the cabin empty:
$P_{p,empty}=(400-150)kg*9.81\frac{m}{s^2}*1.2\frac{m}{s}=2.94kW$
The potential power at full load if no counterweight were used is:
$P_{p,full-nocounterweight}=800kg*9.81\frac{m}{s^2}*1.2\frac{m}{s}=9.42kW$
If there were friction the power to overcome the friction will be:
$P_{friction}=\frac{F_{f}z}{t}=F_{f}V=800N*1.2\frac{m}{s}=0.96kW$
Then the power required with the cabin empty taking in count friction will be:
$P_{p,friction-empty}=P_{p,empty}+P_{friction}=2.94kW+0.96kW=3.90kW$
