A 2.7 × 10
3
kg elevator carries a maximum
load of 888.9 kg. A constant frictional force
of 2.1 × 10
3
N s the elevator’s motion
upward.
The acceleration of gravity is 9.81 m/s
2
.
What minimum power must the motor deliver to lift the fully loaded elevator at a constant speed 2.10 m/s?
Answer in units of k
To find the minimum power that the motor must deliver to lift the fully loaded elevator at a constant speed, we can use the following formula:
Power = Force × Velocity
First, let's calculate the force required to lift the elevator at a constant speed. Since the elevator is moving upward and there is a constant frictional force retarding its motion, the net force required to overcome the friction and lift the elevator is the sum of the weight of the elevator and the frictional force:
Net Force = Weight + Frictional Force
The weight of the elevator is the mass of the elevator multiplied by the acceleration due to gravity:
Weight = Mass × Gravity
Now let's calculate the power required using the formula mentioned above. Note that power is typically measured in watts, but in this case, the answer should be given in kilowatts (kW):
Power = Net Force × Velocity
To convert from watts to kilowatts, divide the result by 1000.
Let's calculate the values step by step:
Step 1: Calculate the weight of the elevator:
Weight = Mass × Gravity
Weight = (2.7 × 10^3 kg) × (9.81 m/s^2)
Step 2: Calculate the net force:
Net Force = Weight + Frictional Force
Net Force = (2.7 × 10^3 kg × 9.81 m/s^2) + (2.1 × 10^3 N)
Step 3: Calculate the power required:
Power = Net Force × Velocity
Power = (Net Force) × (2.10 m/s)
Step 4: Convert power to kilowatts:
Power in kilowatts = Power (in watts) / 1000
Now you can plug in the values into these equations and calculate the minimum power required in units of kilowatts.
To calculate the minimum power required to lift the fully loaded elevator at a constant speed, we can start by calculating the total force acting on the elevator.
The weight of the elevator can be calculated by multiplying the mass by the acceleration due to gravity:
Weight = mass * acceleration due to gravity
Weight = (2.7 * 10^3 kg + 888.9 kg) * 9.81 m/s^2
Next, we need to consider the frictional force acting in the opposite direction. The frictional force can be subtracted from the weight force to find the net force on the elevator:
Net force = Weight - frictional force
Net force = (2.7 * 10^3 kg + 888.9 kg) * 9.81 m/s^2 - 2.1 * 10^3 N
Now we can calculate the minimum power required to overcome the net force and lift the elevator at a constant speed. Power is defined as the product of force and velocity:
Power = Net force * velocity
Power = (2.7 * 10^3 kg + 888.9 kg) * 9.81 m/s^2 - 2.1 * 10^3 N * 2.10 m/s
Finally, let's convert the power to kilowatts (kW):
Power (kW) = Power / 1000
Now we can plug in the values and calculate the minimum power required to lift the fully loaded elevator at a constant speed of 2.10 m/s.
m = m1 + m2 = 2700 + 888.9 = 3589 kg. =
Total mass.
F = mg = 3589 * 9.81 = 35,208 N.
Fn = Fap - F - Fr = 0, a = 0.
Fap - 35208 - 2100 = 0,
Fap = 37,208 N. = Force applied by motor.
Power = F * V = 37,208 * 2 m/s = 74,416
Joules/s = 74,416 Watts. = 74.416 Kilowatts.