Power is needed for (1) acceleration and (2) lifting the loaded chairs. These two parts can be calculated separately and then added together.
(1) Power for acceleration:
The final speed of the lift is
V=(10 km/h)(1 h×1000 m60 sec×1 km)=2.887 m/s.
Then the power needed is
Pa=12m(V2−V20)/Δt=12(50×250 kg)(2.778 m/s)2=9.6 kW.
(2) Power for lift
Assume that the acceleration is constant (i.e. power supply is constant), its value will be
a=ΔVΔt=2.778 m/s5 s=0.556 m/s2.
Then the vertical lift during acceleration will be
(12at2)×(2001000)=1.36 m.
Hence, the power needed to increase the potential energy of the lift is
Pg=mgΔhΔt=(50×250 kg) (9.89 m/s2)(1.36 m)/(5 s)=3.41 kW.
Then the total Power required is
Ptotal=Pa+Pg=9.6+34.1=43.7 kW.
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