An elevator m=800kg has a maximum load of 600kg the elevator goes up 30m at a constant speed of 4ms_¹ what is the average power output of the elevator motor if she elevator is fully loaded with it's maximum m weight (neglect friction)
m = 800 + 600 = 1400 kg
gain of potential energy in (30/4) seconds = m g h = 1400 * 9.81 * 30 Joules
power = energy/time =1400 * 9.81 * 30 / (30/4) = 1400 * 9.81 * 4 Watts
You could also say power = force * speed = 1400 * 9.81 * 4
thank you
To find the average power output of the elevator motor, we need to calculate the work done by the elevator in moving the loaded mass to a height of 30m.
The work done (W) is given by the formula:
W = m * g * h
Where:
m = mass (800kg)
g = acceleration due to gravity (approximately 9.8 m/s²)
h = height (30m)
W = 800kg * 9.8 m/s² * 30m
W = 235,200 Joules
The time taken to cover a distance of 30m at a constant speed of 4 m/s can be calculated using the formula:
time = distance / speed
time = 30m / 4 m/s
time = 7.5 seconds
Average power (P) is given by the formula:
P = W / t
P = 235,200 Joules / 7.5 seconds
P = 31,360 Watts (or 31.36 kilowatts)
Therefore, the average power output of the elevator motor when fully loaded with its maximum weight is 31,360 Watts.