JEE MAIN - Physics (2020 - 7th January Morning Slot - No. 19)
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)
(1 HP = 746 W, g = 10 ms-2)
1.5 ms-1
1.7 ms-1
2.0 ms-1
1.9 ms-1
Explanation
F = mg + f
F = 20000 + 4000 = 24000 N
We know, Power(P) = Fv
$$ \Rightarrow $$ v = $${P \over F}$$ = $${{60 \times 746} \over {24000}}$$
$$ \Rightarrow $$ v $$ \approx $$ 1.9 m/s
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