Question

8. A lift that can carry a maximum load of 2000 kg is moving up with a constant speed of 4 m/s. The frictional force opposing the motion is 5000 N Calculate the minimum power delivered by the motor to elevator. 1.00​

Answer 1

The minimum power delivered by the motor to the elevator is 20,000 Watts.

Step-by-step explanation:

To calculate the minimum power delivered by the motor to the elevator, we need to first determine the force exerted by the motor. We can use Newton's second law of motion, which states that the net force on an object is equal to its mass multiplied by its acceleration. In this case, the net force is the sum of the force exerted by the motor and the frictional force opposing motion. So, we have:

Net Force = Force by motor - Frictional force

The acceleration of the elevator is zero because it is moving at a constant speed. Therefore,

Force by motor = Frictional force

Now, we can calculate the power delivered by the motor using the equation:

Power = Force by motor x speed

Substituting the given values, we have:

Power = 5000 N x 4 m/s = 20000 W

Therefore, the minimum power delivered by the motor to the elevator is 20,000 Watts.

Answer 2

Explanation:To calculate the minimum power delivered by the motor to the elevator, you can use the following formula for power:

Power (P) = Force (F) × Velocity (v)

In this case, you're interested in the power required to overcome friction while maintaining a constant speed. The force opposing the motion is the frictional force, which is given as 5000 N, and the velocity is 4 m/s.

P = 5000 N × 4 m/s

P = 20,000 N·m/s (Watts)

So, the minimum power delivered by the motor to the elevator is 20,000 Watts, which is equivalent to 20,000 joules per second or 20 kilowatts (kW).