A 1000kg elevator carries a maximum load of 800kg. A constant frictional force of 5000N s the elevator's motion upward. What minimum power, in kilowatts, must the motor deliver to lift the fully loaded elevator at a constant speed of 3m/s?
To find the minimum power the motor must deliver, we can use the equation:
Power = Force x Velocity
First, let's calculate the net force acting on the elevator. The force can be determined by considering the difference between the weight of the elevator and the maximum load it carries:
Weight of the elevator (mg) = 1000kg x 9.8m/s^2 = 9800N
Maximum load carried = 800kg x 9.8m/s^2 = 7840N
Net force = Weight of the elevator - Maximum load carried = 9800N - 7840N = 1960N
Since the motion is upward, and there is a constant frictional force of 5000N retarding the motion, the minimum force required to lift the elevator at a constant speed would be:
Minimum force = Net force + Frictional force = 1960N + 5000N = 6960N
Now, let's calculate the power using the formula:
Power = Force x Velocity
Power = 6960N x 3m/s = 20880 Nm/s
Finally, to convert the power into kilowatts, we divide by 1000:
Power = 20880 Nm/s / 1000 = 20.88 kW
Therefore, the minimum power the motor must deliver to lift the fully loaded elevator at a constant speed of 3m/s is 20.88 kilowatts.
To calculate the minimum power required for the elevator, we need to consider both the force of gravity acting on the elevator as well as the work done against friction.
First, let's calculate the force of gravity acting on the elevator. The force of gravity is given by the formula:
Force_gravity = mass * gravitational acceleration
where the mass of the elevator is the sum of the maximum load and the mass of the elevator itself. The gravitational acceleration is approximately 9.8 m/s².
mass_elevator = 1000 kg + 800 kg = 1800 kg
Force_gravity = 1800 kg * 9.8 m/s² = 17,640 N
Next, we need to calculate the net force required to lift the elevator at a constant speed of 3 m/s, taking into account the frictional force. Since the elevator is moving at a constant speed, the net force must be zero.
net_force = Force_up - Force_friction
Since both the force of gravity and the frictional force are acting in the opposite direction of the elevator's motion, we have:
net_force = Force_gravity - Force_friction
0 = 17,640 N - 5000 N
Force_up = 5000 N
Finally, we can calculate the minimum power required using the formula:
Power = force * velocity
where both force and velocity are in the same direction.
Power = 5000 N * 3 m/s = 15,000 N·m/s
To convert the power to kilowatts, we divide by 1000:
Power = 15,000 N·m/s ÷ 1000 = 15 kW
Therefore, the minimum power, in kilowatts, that the motor must deliver to lift the fully loaded elevator at a constant speed of 3 m/s is 15 kW.