An electric lift of total mass 2000 kg moves upwards at constant 1.5 m s⁻¹ against a friction force of 3000 N. Take g=10 m s⁻². Minimum motor power is:
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Question

An electric lift of total mass 2000 kg moves upwards at constant 1.5 m s⁻¹ against a friction force of 3000 N. Take g=10 m s⁻². Minimum motor power is:

G1 Physics · Accepted Answer

Step 1 — Principle: Instantaneous power from a force is P = F·v. Step 2 — Working: At constant speed upward motor force=mg+f=20000+3000=23000 N. Step 3 — Working: Power P=Fv=23000×1.5=34500 W. Correct answer: B — 34500 W.