Question

An elevator whc can carry a max load of 1800 kg is moving up with a constant speed of 2m/s. the frictional force opposing motion is 4000n. determine the min power delivered by the motor to the elevator

Answer 1

I don't know the answer of ur question sorry for that

Answer 2

Answer :

➟ P = 43280 watts

Explanation :

Given :

➟ Mass of load, m = 1800 kg

➟ Speed of elevator, v = 2 m/s

➟ Frictional force, f = 4000 N

To find :-

➟ Min power delivered by the motor to the elevator

Solution :-

$\sf {\implies}\:F_{net}=mg+f$

$\sf {\implies}\:F_{net}=1800\times 9.8+4000$

$\sf {\implies}\:F_{net}=21640\ N$

$\rule {300}{1.5}$

We have to find the minimum power delivered by the motor to the elevator.

$\sf {\implies}\:P=F_{net}\times v$

$\sf {\implies}\:P=21640\times 2$

➟ P = 43280 Watts