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An elevator can carry a maximum load of `1800 kg` (elevator + passengers) is moving up with a constant speed of `2 ms^(-1)`. The friction force opposite the motion is `4000 N`.What is minimum power delivered by the motor to the elevator?A. `59`B. `8`C. `22`D. `20` |
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Answer» Correct Answer - A `F = mg + "frictional force"`, `P = vec(F).vec(v)` |
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