JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
A \(60\; HP\) electric motor lifts an elevator having a maximum total load capacity of \(2000\; \mathrm{kg}\). If the frictional force on the elevator is \(4000 \;\mathrm{N}\). the speed of the elevator at full load is close to .............. \(\mathrm{m} / \mathrm{s}\) \(\left(1 \;\mathrm{HP}=746 \;\mathrm{W}, \mathrm{g}=10\; \mathrm{ms}^{-2}\right)\)
- A \(1.7\)
- B \(2\)
- C \(1.9\)
- D \(1.5\)
Answer & Solution
Correct Answer
(C) \(1.9\)
Step-by-step Solution
Detailed explanation
Let elevator is moving upward with constant speed V. Tension in cable \(\mathrm{T}=2000 \mathrm{g}+\mathrm{f}_{\mathrm{r}}=2000+4000\) \(\mathrm{T}=24000 \mathrm{N}\) 10.Power \(P=T V\) \(\Rightarrow 60 \times 746=(24000) \mathrm{V}\)…
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