Q.An elevator can carry a maximum load of $1800\ \text{kg}$ (elevator + passengers) is moving up with a constant speed of $2\ \text{m s}^{-1}$. The frictional force opposing the motion is $4000\ \text{N}$. Determine the minimum power delivered by the motor to the elevator in watts as well as in horse power.

The key idea is that the power delivered by the motor must overcome the forces opposing the elevator's upward motion: gravity and friction. Since the elevator moves at a constant speed, the net force is zero, meaning the motor's upward force equals the sum of the downward forces.

1. The total upward force required from the motor ($F_m$) must balance the gravitational force ($mg$) and the frictional force ($F_f$).

   $F_m = mg + F_f$

   $F_m = (1800\ \text{kg})(10\ \text{m s}^{-2}) + 4000\ \text{N}$

   $F_m = 18000\ \text{N} + 4000\ \text{N} = 22000\ \text{N}$

2. The power ($P$) delivered by the motor is the product of this force and the constant speed ($v$).

   $P = F_m \cdot v$

   $P = (22000\ \text{N})(2\ \text{m s}^{-1}) = 44000\ \text{W}$ …

When an elevator moves upwards at a constant speed, the motor isn't accelerating the elevator; rather, it's doing work against the forces that oppose its upward motion. These opposing forces are primarily gravity, pulling the elevator and its load downwards, and frictional forces, which resist any motion. The "minimum power" implies we are looking for the power required to just maintain this constant speed, meaning there is no net acceleration.

The concept of power here is the rate at which the motor does work to overcome these resistive forces. If a force $F$ is applied to move an object at a constant velocity $v$ in the direction of the force, the power $P$ delivered is given by the product of the force and the velocity.

In this problem, the motor's upward force must exactly balance the sum of the downward gravitational force and the downward frictional force. Since the velocity is constant, the net force on the elevator is zero.

1. Identify the forces opposing upward motion:

   • Gravitational force ($F_g$): This is the weight of the elevator plus its maximum load. It acts downwards.
   • Frictional force ($F_f$): This force opposes the motion, so for upward motion, it acts downwards.

2. Calculate the gravitational force:

   The total mass ($m$) of the elevator and passengers is $1800\ \text{kg}$. We use $g = 10\ \text{m s}^{-2}$, the value this chapter's worked examples use throughout (e.g. Example 5.2).

3. Determine the total upward force required from the motor: Since the elevator is moving at a constant speed, its acceleration is zero. According to Newton's second law, the net force on the elevator must be zero. This means the upward force exerted by the motor ($F_{\text{motor}}$) must exactly balance the sum of the downward gravitational force and the downward frictional force.

We are given the frictional force $F_f = 4000\ \text{N}$.

This is the minimum force the motor must exert to maintain constant upward speed. …

Concept: Power Delivered Against Gravity and Friction (Constant Velocity)

Step 1: Find the total upward force the motor must supply.

At constant speed the net force is zero, so the motor's force balances gravity plus friction (using $g = 10\,\text{m/s}^2$, as this chapter's worked examples do throughout):

Step 2: Compute the power using $P = F\cdot v$. …