A(n) 636 kg elevator starts from rest. It moves upward for 4.5 s with a constant acceleration until it reaches its cruising speed of 2.05 m/s. The acceleration of gravity is 9.8 m/s 2 . Find the average power delivered by the elevator motor during this period. Answer in units of kW.

The average power delivered by the elevator motor during this period is 6.686 kW.

Explanation:

Given;

mass of the elevator, m = 636 kg

initial speed of the elevator, u = 0

time of motion, t = 4.5 s

final speed of the elevator, v = 2.05 m/s

The upward force of the elevator is calculated as;

F = m(a + g)

where;

m is mass of the elevator

a is the constant acceleration of the elevator

g is acceleration due to gravity = 9.8 m/s²

F = (636)(0.456 + 9.8)

F = (636)(10.256)

F = 6522.816 N

The average power delivered by the elevator is calculated as;

Therefore, the average power delivered by the elevator motor during this period is 6.686 kW.

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pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y