An elavater m=800kg has a maximum load of 8 people of 600kg. The elevatorgoes up to 10 stories = 30m at a costant speed of 4m/s. what is the average power output of the elevater motor if the elevater is fully loaded with its maximum weight? (neglect friction)
power= work/time= (600+800)g*30/(time)
where time is 30/4 seconds
P = F v = w v = (800+600)(4) = 5600 W
To calculate the average power output of the elevator motor, we need to find the work done by the elevator and the time it takes to complete that work.
First, let's calculate the work done by the elevator. Work is defined as the product of force and displacement. In this case, the force exerted by the elevator is equal to its mass multiplied by the gravitational acceleration (F = mg). The displacement is the height the elevator moves, which is 30m.
The force exerted by the elevator is given by F = (800 kg + 8 people x 75 kg/person) x 9.8 m/s² (assuming standard gravitational acceleration of 9.8 m/s²).
F = (800 kg + 8 x 75 kg) x 9.8 m/s²
F = 800 kg + 600 kg x 9.8 m/s²
F = 800 kg + 5880 kg m/s²
F = 6680 kg m/s²
The work done by the elevator is given by W = Fs, where s is the displacement of 30m.
W = (6680 kg m/s²) x 30m
W = 200,400 kg m²/s²
Now, let's find the time it takes for the elevator to complete this work. Time is given by t = s/v, where s is the displacement and v is the velocity.
t = 30m / 4m/s
t = 7.5 seconds
Finally, we can calculate the average power output using the formula P = W/t, where P is the average power output, W is the work done, and t is the time taken.
P = (200,400 kg m²/s²) / 7.5 seconds
P ≈ 26,720 W (rounded to the nearest whole number)
Therefore, the average power output of the elevator motor when fully loaded with its maximum weight is approximately 26,720 watts.