An elevator lifts a load of 500 kg with a co | vedclass

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(C) The power $P$ required to lift a load at a constant speed is given by the formula $P = Fv$,where $F$ is the force and $v$ is the velocity.Since th

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$2.62$

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(C) The power $P$ required to lift a load at a constant speed is given by the formula $P = Fv$,where $F$ is the force and $v$ is the velocity.Since the elevator moves at a constant speed,the force $F$ is equal to the weight of the load,$mg$.Given: $m = 500 \ kg$,$g = 9.8 \ m/s^2$,$v = 0.4 \ m/s$.$P = (mg)v = 500 	imes 9.8 	imes 0.4 = 1960 \ W$.To convert power from Watts to Horsepower $(H.P.)$,we use the conversion factor $1 \ H.P. = 746 \ W$.$P = \frac{1960}{746} \approx 2.62 \ H.P.$

An elevator lifts a load of $500 \ kg$ with a constant speed of $0.4 \ m/s$. What is the minimum horsepower $(H.P.)$ of the motor required?

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