Answer:
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
[tex]a = \frac{v-u}{t} \\\\a = \frac{2.05 -0}{4.5} \\\\a = 0.456 \ m/s^2[/tex]
F = (636)(0.456 + 9.8)
F = (636)(10.256)
F = 6522.816 N
The average power delivered by the elevator is calculated as;
[tex]P_{avg} = \frac{1}{2} (Fv)\\\\P_{avg} = \frac{1}{2} (6522.816 \ \times \ 2.05)\\\\P_{avg} = 6685.89 \ W\\\\P_{avg} = 6.68589 \ kW\\\\P_{avg} = 6.686 \ k W[/tex]
Therefore, the average power delivered by the elevator motor during this period is 6.686 kW.
