Question 139: Work, Energy and Power & Circular Motion

Question

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
)

Accepted Answer

F = mg + f
F = 20000 + 4000 = 24000 N
We know, Power(P) = Fv
$$ \Rightarrow $$ v = $${P \over F}$$ = $${{60 	imes 746} \over {24000}}$$
$$ \Rightarrow $$ v $$ \approx $$ 1.9 m/s

Answer

1.5 ms
-1

Answer

1.7 ms
-1

Answer

2.0 ms
-1

Answer

1.9 ms
-1

Work, Energy and Power & Circular Motion

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 )

Choose the correct answer: