A small domestic elevator has a mass of 454 kg and can ascend at a rate of 0.180 m/s. What is the average power that must be supplied for the elevator to move at this rate ? answer in watts.
Fe = 454kg * 9.8N/kg = 4449.2N.
Pe Fd/t = FV = 4449.2 * 0.180m/s = 801W
To calculate the average power required for the elevator to move at a certain rate, we can use the equation:
Power (P) = Force (F) x Velocity (v)
First, let's calculate the force:
Force (F) = Mass (m) x Acceleration (a)
Since the elevator is ascending at a constant rate, the acceleration is 0. Therefore, the force will be:
Force (F) = Mass (m) x Acceleration (a) = 454 kg x 0 = 0 N
Now, we can calculate the power:
Power (P) = Force (F) x Velocity (v) = 0 N x 0.180 m/s = 0 W
Therefore, the average power required for the elevator to move at this rate is 0 watts.
To find the average power required for the elevator to move at this rate, we can use the formula:
Power = Force × Velocity
First, we need to calculate the force that the elevator needs to overcome to move upwards. This force is equal to the weight of the elevator, which can be calculated using the equation:
Weight = mass × acceleration due to gravity
The acceleration due to gravity is approximately 9.8 m/s². So, the weight of the elevator is:
Weight = 454 kg × 9.8 m/s² = 4451.2 N
Next, we need to calculate the work done by the elevator. Work is equal to the force multiplied by the displacement:
Work = Force × Distance
Since the elevator is ascending at a constant velocity, the distance traveled is equal to the velocity multiplied by the time:
Distance = Velocity × Time
Now, we can substitute the values into the equation:
Work = 4451.2 N × (0.180 m/s × 1 s) = 801.216 J
Finally, we can calculate the power:
Power = Work / Time
Since the work was calculated over a period of 1 second, the power is equal to the work:
Power = 801.216 J
So, the average power required for the elevator to move at this rate is approximately 801.216 watts.