First slide

Power

Question

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 4000N, the speed of the elevator at full load is close to: $(1 H P = 746 W, g = 10 m s^{- 2})$

Easy

A

1.9m/s
B

1.7m/s
C

2.0m/s
D

1.5m/s

Solution

$\begin{array}{l} B o t h m g a n d f r i c t i o n a r e a c t i n g d o w n w a r d s . \\ I f e l e v a t o r i s m o v i n g u p w i t h u n i f o r m v e l o c i t y, \\ t h e n \\ F_{f r} V + F_{m g} V = P \\ \Rightarrow 4000 \times V + (2000 \times 10) \times V = (60 \times 746) \\ \Rightarrow V = \frac{60 \times 746}{4000 + 20000} \\ \Rightarrow V = 1.865 m / s \approx 1.9 m / s \end{array}$

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