A(n) 563 kg elevator starts from rest. It moves upward for 3.18 s with a constant acceleration until it reaches its cruising speed of 1.96 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 elevator accelerates at a = 1.96/3.18 = 0.6134 m/s^2

The height increase during the acceleration period is
H = (1/2) a t^2 = 3.10 m

The energy gain during that interval is
M*g*H + (M/2)*V^2

Divide by the time interval for the average power

To find the average power delivered by the elevator motor, we need to use the formula:

Average Power = (Change in Energy) / (Time)

First, let's find the change in energy of the elevator. We know that the elevator starts from rest, so its initial velocity (v₀) is 0 m/s. The final velocity (v) of the elevator is 1.96 m/s.

Using the kinematic equation:

v = v₀ + at

where:
v = final velocity = 1.96 m/s
v₀ = initial velocity = 0 m/s
a = acceleration
t = time = 3.18 s

Rearranging the equation, we can solve for the acceleration:

a = (v - v₀) / t
= (1.96 m/s - 0 m/s) / 3.18 s
≈ 0.615 m/s²

Now, let's find the change in energy. The change in kinetic energy is given by:

ΔKE = (1/2) * m * (v² - v₀²)

where:
ΔKE = change in kinetic energy
m = mass of the elevator = 563 kg
v = final velocity = 1.96 m/s
v₀ = initial velocity = 0 m/s

Substituting the values into the equation:

ΔKE = (1/2) * 563 kg * (1.96 m/s)²

Calculating ΔKE:

ΔKE ≈ 1099.31 J

Now, let's calculate the average power delivered by the elevator motor using the formula:

Average Power = ΔKE / t

where:
Average Power = average power delivered by the motor
ΔKE = change in kinetic energy = 1099.31 J
t = time = 3.18 s

Substituting the values into the equation:

Average Power = 1099.31 J / 3.18 s

Calculating the average power:

Average Power ≈ 345.08 W

Finally, let's convert the average power from watts to kilowatts:

Average Power = 345.08 W / 1000

Calculating the average power in kilowatts:

Average Power ≈ 0.345 kW

Therefore, the average power delivered by the elevator motor during this period is approximately 0.345 kW.