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)
Solution
<br><br>F = mg + f
<br><br>F = 20000 + 4000 = 24000 N
<br><br>We know, Power(P) = Fv
<br><br>$\Rightarrow$ v = ${P \over F}$ = ${{60 \times 746} \over {24000}}$
<br><br>$\Rightarrow$ v $\approx$ 1.9 m/s
About this question
Subject: Physics · Chapter: Work, Energy and Power · Topic: Work Done by a Force
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