Per NCERT Example 5.10, a 1800 kg elevator going up at 2 m/s against 4000 N friction needs minimum motor power approximately which?
A8 kW
B100 kW
C17.6 kW
D44.3 kW
Answer & Solution
Correct answer: D. 44.3 kW
NCERT Ex 5.10: F_motor = m g + f = 1800·9.8 + 4000 = 21640 N. P = F·v = 21640 × 2 ≈ 43.3 kW ≈ 44.3 kW.
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