Question

Elevator carry maximum load 1800 kg moves with a constant speed 2m/s . The frictional force opposing this motion is 4000N determine the minimum power delivered by the motor to the elevator in watts and horse power ?

Answer 1

power=work(energy)/time
power=F*s/t.
(F is force, t is time, s is displacement)
power=F*v. (s/t=v or velocity)
power=(m*g+4000)*2.
(m is mass g is acceleration due to gravity)
mg is the force because it the minimum earth pull and 4000N to overcome the friction
(g=10m/s²)
power=(1800*10+4000)*2
power=(18000+4000)*2
power=22000*2
power=44000 watt