An elevator which can carry a maximum load of 1800 kg(elevator + passengers) is moving up with a constant speed of 2m/s.The frictional force opposing the motion is 4000 N.Determine the minimum power delivered by the motor to the elevator in watts as well as in horse power.
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2 Answers
I think Power done by motor is the work done on elevator + work done against frictional forces.
So, Wm = K.E of elevator + Wf
I think you can find it and then differentiate the both sides setting the RHS to 0. You will find a certain value of the variable and you put it back into the expression..
Don't worry. TRY!!
since force of gravity and friction will be constant , u can simply use power = F . v
since elevator is moving with const speed , motor balances the forces of gravity and friction on the elevator
hence F = 18000 + 4000= 22000 N
hence power = F . v = 22000 x 2 x cos 0 = 44000 W