Question

An elevator can carry a maximum load of 1800kg, is moving up with a constant speed of 2m/s. The frictional force opposing the motion is 4000N. Determine the maximum power delivered by the motor to the elevator in watt as well as in horse power.

Answer 1

Answer:

59hp

Explanation:

hence frictional force and force of motion is acting downward

$(net{force}) = mg + f \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = (1800kg)(10) + 4000 \: \: \: \: \: \: \: \: \: \: \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 22000$

$power = 22000 \times 2ms ^{ - }{1 } \: \: \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 44000watt$

hence power of the motor =44000watt =59hp

Answer 2

Explanation:

$\Large{\red{\underline{\underline{\sf{\blue{Given:}}}}}}$

$\sf Mass,\:m\:=\:1800kg$

$\sf Frictional\:force,\:f\:=\:4000N$

$\sf Uniform\:speed,\:v\:=\:2.0ms^{-1}$

 ‎ ‎ ‎ ‎ ‎ ‎ ‎

$\Large{\blue{\underline{\underline{\sf{\red{To\:find:}}}}}}$

$\sf Power\:(in\:watt\:and\:horse\:power)$

 ‎ ‎ ‎ ‎ ‎ ‎

$\Large{\red{\underline{\underline{\sf{\pink{Solution:}}}}}}$

$\sf Downward\:force\:on\:elevator\:is$

 ‎ ‎ ‎ ‎ ‎ ‎

$\sf F\:=\:mg+f$

 ‎ ‎ ‎ ‎ ‎ ‎

$\sf \quad=\:1800\times 10+4000=22000N$

 ‎ ‎ ‎ ‎ ‎ ‎

$\sf The\:motor\:must\:supply\:enough\:force\:to\:balance\:this\:force\:as$

$\sf P\:=\:F\times v$

 ‎ ‎ ‎ ‎ ‎ ‎

$\sf \therefore P\:=\:22000 \times 2\:=\:44000\:watt$

 ‎ ‎ ‎ ‎ ‎ ‎

$\sf P\:=\:\dfrac{44000}{746}h.p\:=\:59h.p$