A 650kg elevator starts from rest. It moves upward for 3s with constant acceleration until it reases its cruising speed, 1.75m/s. (a) what is the average power of the elevator motor during this period? (b) How does this compare with its power during an upward cruise with constant speed?
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To find the average power of the elevator motor during the period of acceleration, we need to calculate the work done and divide it by the time taken.
(a) First, let's find the final velocity of the elevator after the acceleration period. We know the initial velocity (0 m/s), the time duration (3 s), and the cruising speed (1.75 m/s). We can use the equation of motion:
v = u + at
where v is the final velocity, u is the initial velocity, a is acceleration, and t is the time duration. Rearranging this equation to solve for acceleration:
a = (v - u) / t
Substituting the given values:
a = (1.75 m/s - 0 m/s) / 3 s
a = 0.583 m/s^2
Now, let's calculate the distance covered during the acceleration period using the equation:
s = ut + (1/2)at^2
where s is the distance, u is the initial velocity, a is acceleration, and t is the time duration. Substituting the values:
s = 0 m/s * 3 s + (1/2)(0.583 m/s^2)(3 s)^2
s = 0 m + (1/2)(0.583 m/s^2)(9 s^2)
s = 0.583 m
Next, we can calculate the work done during the acceleration period using the equation:
work = force * distance
where work is the work done, force is the net force acting on the elevator, and distance is the distance covered. We can determine the net force by using Newton's second law:
force = mass * acceleration
where force is the net force, mass is the mass of the elevator, and acceleration is the acceleration. Substituting the values:
force = 650 kg * 0.583 m/s^2
force = 378.95 N
Finally, we can calculate the work done:
work = force * distance
work = 378.95 N * 0.583 m
work = 220.97 J
The average power is the work done divided by the time taken:
power = work / time
power = 220.97 J / 3 s
power ≈ 73.66 W
The average power of the elevator motor during the acceleration period is approximately 73.66 Watts.
(b) During the upward cruise with constant speed, the elevator is moving at a constant velocity, so the net force and the acceleration are both zero. Therefore, the work done is also zero. In this case, the power is equal to zero as well. So, the power during the upward cruise with constant speed is zero.
In conclusion, the average power of the elevator motor during the period of acceleration is approximately 73.66 Watts, while the power during the upward cruise with constant speed is zero.