An elevator carrying a maximum load of 1800 kg (elevator + passengers)

1.	An elevator carrying a maximum load of 1800 kg (elevator + passengers) is movingup with a constant speed of 2 ms^-1. The frictional force opposing the motion is4000 N. Determine the minimum power delivered by the motor to the elevator.(Take g = 10 m/s^2)
Answer» Answer: 43280 Watt Explanation: elevator MOVES UPWARD HENCE frictional force OPPOSE their motion in downward. so, net force act on elevator =mg+F because mg is act downward and F (frictional force) act also downward. Fnet=(1800 x 9.8+4000) N =(17640+4000)N=21640N now power =Fnet x speed =21640 x 2=43280 watt

An elevator carrying a maximum load of 1800 kg (elevator + passengers) is movingup with a constant speed of 2 ms^-1. The frictional force opposing the motion is4000 N. Determine the minimum power delivered by the motor to the elevator.(Take g = 10 m/s^2)

Answer»

Answer:

43280 Watt

Explanation:

elevator MOVES UPWARD HENCE frictional force OPPOSE their motion in downward.

so, net force act on elevator =mg+F

because mg is act downward and F (frictional force) act also downward.

Fnet=(1800 x 9.8+4000) N

=(17640+4000)N=21640N

now

power =Fnet x speed

=21640 x 2=43280 watt

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An elevator carrying a maximum load of 1800 kg (elevator + passengers) is movingup with a constant speed of 2 ms^-1. The frictional force opposing the motion is4000 N. Determine the minimum power delivered by the motor to the elevator.(Take g = 10 m/s^2)​

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An elevator carrying a maximum load of 1800 kg (elevator + passengers) is movingup with a constant speed of 2 ms^-1. The frictional force opposing the motion is4000 N. Determine the minimum power delivered by the motor to the elevator.(Take g = 10 m/s^2)