A 60 HP electric motor lifts an elevator having a maximum total load capacity of 2000 kg. If the frictional force on the elevator is 4000 N, the speed of the elevator at full load is close to: (1 HP = 746 W, g = 10 ms –2 )
Text Solution
Verified by ExpertsThe correct answer is:
B
4000 × V + mg × V = P
= V
V = 1.86 m/s. 
m/s.
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