An elevator car (see figure (a)) has a mass of 1,460 kg and is carrying passengers having a combined mass of 220 kg. Motor A constant friction force of 4,050 N retards its motion. (A) How much power must a motor deliver to lift the elevator car and its passengers at a constant speed of 2.20 m/s? (B) What power must the motor deliver at the instant the speed of the elevator is v if the motor is designed to provide the elevator car with an upward acceleration of 1.80 m/s?? SOLVE IT (A) How much power must a motor deliver to lift the elevator car and its passengers at a constant speed of 2.20 m/s? Conceptualize The motor must supply the force of magnitude T that pulls the elevator car upward. Categorize The friction force increases the power necessary to lift the elevator. The problem states that the speed of the elevator is constant, which tells us that a = 0. We model the elevator as a particle in equilibrium. (a) The motor exerts an upward force T on the elevator car. The magnitude of this force is the total tension T in the cables connecting the car and motor. The downward forces acting on the car are a friction force f and the gravitational force F- ME. (b) The free-body %3D Analyze The free-body diagram in figure (b) specifies the upward direction as positive. The total mass M of the gers) is equal to 1,680 kg. diagram for the elevator car. system (car plus pass Using the particle in equilibrium model, apply Newton's second law to the car: E, =T-1- Mg = 0 T = Mg + f Solve for T: Use P = F. V and that Tis in the same direction as v to find the power: P = T.7 = TV = (Mg + nv Substitute numerical values: P = [(1,680 kg)(9.80 m/s?) + (4,050 N)](2.20 m/s) x W (B) What power must the motor deliver at the instant the speed of the elevator is v if the motor is designed to provide the elevator car with an upward acceleration of 1.80 m/s? Conceptualize In this case, the motor must supply the force of magnitude T that pulls the elevator car upward with an increasing speed. We expect that more power will be required to do that than in part (A) because the motor must now perform the additional task of accelerating the car. Categorize In this case, we model the elevator car as a particle under a net force because it is

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**Transcription for Educational Website**

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### Problem Statement

An elevator car (see figure (a)) has a mass of 1,460 kg and carries passengers with a combined mass of 220 kg. A constant friction force of 4,050 N opposes its motion.

**(A)** How much power must a motor deliver to maintain the elevator car and its passengers at a constant speed of 2.20 m/s?

**(B)** What power must the motor deliver at the instant when the elevator speed is \( v \), if the motor is designed to provide the elevator car with an upward acceleration of 1.80 m/s\(^2\)?

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### Solution

#### Part (A)

**Conceptualize:** The motor must supply an upward force \( T \) to move the elevator car.

**Categorize:** The friction force increases the power needed to lift the elevator. Given that the elevator speed is constant, we conclude that acceleration \( a = 0 \). The elevator is treated as a particle in equilibrium.

**Analyze:** 
- The free-body diagram in figure (b) sets the upward direction as positive.
- Total mass \( M \) of the system (car plus passengers) is 1,680 kg.

Apply Newton's second law to the car:
\[
\sum F_y = T - f - Mg = 0
\]

Solve for \( T \):
\[
T = Mg + f
\]

To find the power, \( P = \mathbf{F} \cdot \mathbf{v} \) where \( \mathbf{T} \) and \( \mathbf{v} \) are in the same direction:
\[
P = T \cdot v = (Mg + f)v
\]

Substitute numerical values:
\[
P = [(1,680 \, \text{kg})(9.80 \, \text{m/s}^2) + (4,050 \, \text{N})](2.20 \, \text{m/s})
\]

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#### Part (B)

**Conceptualize:** For this scenario, the motor must provide the force \( T \) to accelerate the elevator car upwards. More power is required compared to part (A) due to the need for acceleration.

**Categorize:** The elevator car is considered a particle under a net force because it accelerates.

**Analyze:** Apply Newton's second law
Transcribed Image Text:**Transcription for Educational Website** --- ### Problem Statement An elevator car (see figure (a)) has a mass of 1,460 kg and carries passengers with a combined mass of 220 kg. A constant friction force of 4,050 N opposes its motion. **(A)** How much power must a motor deliver to maintain the elevator car and its passengers at a constant speed of 2.20 m/s? **(B)** What power must the motor deliver at the instant when the elevator speed is \( v \), if the motor is designed to provide the elevator car with an upward acceleration of 1.80 m/s\(^2\)? --- ### Solution #### Part (A) **Conceptualize:** The motor must supply an upward force \( T \) to move the elevator car. **Categorize:** The friction force increases the power needed to lift the elevator. Given that the elevator speed is constant, we conclude that acceleration \( a = 0 \). The elevator is treated as a particle in equilibrium. **Analyze:** - The free-body diagram in figure (b) sets the upward direction as positive. - Total mass \( M \) of the system (car plus passengers) is 1,680 kg. Apply Newton's second law to the car: \[ \sum F_y = T - f - Mg = 0 \] Solve for \( T \): \[ T = Mg + f \] To find the power, \( P = \mathbf{F} \cdot \mathbf{v} \) where \( \mathbf{T} \) and \( \mathbf{v} \) are in the same direction: \[ P = T \cdot v = (Mg + f)v \] Substitute numerical values: \[ P = [(1,680 \, \text{kg})(9.80 \, \text{m/s}^2) + (4,050 \, \text{N})](2.20 \, \text{m/s}) \] --- #### Part (B) **Conceptualize:** For this scenario, the motor must provide the force \( T \) to accelerate the elevator car upwards. More power is required compared to part (A) due to the need for acceleration. **Categorize:** The elevator car is considered a particle under a net force because it accelerates. **Analyze:** Apply Newton's second law
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