Required power for equally balanced elevator cab-counterweight

[suratpanas]

Homework Statement

An elevator system (traction system) consists of a cab, a counterweight, and the motor-powered pulley. The cab and the counterweight is connected by a rope which is attached on the pulley. Consider the cab and the counterweight have equal mass = 1000 kg.
The elevator is intended to move at 1 m/s. This is a balanced mass on 2 sides of the pulley. Calculate the motor power required.

mCa = mCo = 1000 kg
v = 1 m/s

P (kW) = ?

Homework Equations

F dt = mv
P = F * x/t

The Attempt at a Solution

The elevator is initially at rest, so initial momentum is simply zero. No information about the acceleration, so let's say the cruising speed of 1 m/s is reached in 1 s.
F = (mCa + mCo) * v / t
F = 2000 * 1 / t
say t = 1
F = 2000

The distance required to accelerate to cruising speed is
x = 0.5 * a * t^2
since we have stated that 1 m/s reached in 1 sec, then a = 1 m/s^2. Thus,
x = 0.5 * 1 * 1^2 = 0.5

The power calculation:
P = F * x / t
P = 2000 * 0.5/1
P = 1000 W = 1 kW

Conclusion:

1. When accelerating to it's cruising speed (1 m/s), the system requires 1 kW from the motor regardless the friction.

2. when it's already in its cruising speed, the power required is ZERO, regardless the friction. Means, the power is only to overcome the friction.


Am I right, folks? did I do correctly in my analysis of using momentum? This is equally (balanced) masses at both sides, so gravitational forces cancel each other.
Thanks for your comments..

 

[anathema]

Thank you for your detailed explanation and calculations. Your analysis is correct in terms of using momentum to calculate the force required for the elevator to reach its cruising speed of 1 m/s. However, there are a few additional factors to consider in determining the motor power required for this elevator system.

Firstly, as you mentioned, there will be friction present in the system that will require additional force to overcome. This friction can come from various sources such as the pulley, the rope, and even air resistance. Therefore, the force required to move the elevator will be greater than just the force calculated from the momentum equation.

Secondly, the elevator system will also have to overcome the weight of the cab and counterweight, which is equal to 2000 kg in this case. This weight will also require additional force from the motor to lift and move the elevator.

Taking these factors into account, the motor power required for this elevator system will be greater than just 1 kW. It would be necessary to calculate the total force required, taking into account the weight of the system and the friction, and then use that value to calculate the motor power using the equation P = F * v.

I hope this helps clarify your analysis. Keep up the good work in your scientific calculations!