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A cable lifts a 1200 -kg elevator at a constant velocity for a distance of \(35 \mathrm{m} .\) What is the work done by (a) the tension in the cable and (b) the elevator's weight?

Short Answer

Expert verified
The work done by the tension is 411,600 J and by the elevator's weight is -411,600 J.

Step by step solution

01

Understanding Work Done by Forces

Work is given by the formula \( W = F \cdot d \cdot \cos(\theta) \), where \( F \) is the force acting on the object, \( d \) is the distance over which the force is applied, and \( \theta \) is the angle between the force and the direction of motion. For vertical motion at constant velocity, the work done by tension in the cable will equal the work done against the elevator's weight, with the angle being 0 degrees, meaning \( \cos(0) = 1 \).
02

Calculating Work Done by Tension

First, calculate the force of tension in the cable. Since the elevator moves at constant velocity, tension equals the weight of the elevator. The gravitational force \( F_g \) on the elevator is calculated as:\[ F_g = m \cdot g = 1200 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 11760 \, \text{N} \]Since \( \theta = 0 \), work done by tension \( W_T \) is:\[ W_T = F_T \cdot d \cdot \cos(0) = 11760 \, \text{N} \times 35 \, \text{m} = 411600 \, \text{J} \]
03

Calculating Work Done against Elevator's Weight

The work done against the elevator's weight is equal in magnitude but opposite in direction to the work done by the tension. Thus, the work done against the elevator's weight is:\[ W_g = -F_g \cdot d = -11760 \, \text{N} \times 35 \, \text{m} = -411600 \, \text{J} \]
04

Summarizing Work Done

The work done by the tension in the cable is positive, as energy is being transferred to lift the elevator upwards. In contrast, the work done against the gravity force on the elevator is negative, corresponding to the work needed to counteract gravitational pull.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Tension in Cable
When we talk about tension in a cable, especially in the context of lifting heavy objects like elevators, it's essential to understand that tension is the force transmitted through the cable when it is pulled tight by forces acting from opposite ends. This force is crucial because it is responsible for lifting the elevator and moving it upwards.
The tension force in the cable is what counteracts the gravitational force pulling the elevator down. When lifting an object at constant velocity, these two forces are in equilibrium, meaning they are equal in magnitude. This equilibrium ensures a smooth ascent or descent without acceleration.
In the elevator scenario, the tension in the cable is equal to the weight of the elevator when it's moving at a constant velocity. Thus, the formula for tension is:
  • \[ F_T = m imes g \]
where \( m \) represents the mass of the elevator and \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)). Understanding this relationship helps us calculate how much work the tension in the cable performs during the elevator ride.
Elevator Weight
The weight of the elevator is a critical factor in understanding how much force is needed to move it. Weight is the force exerted by gravity on an object and is calculated by the equation \( F_g = m \times g \), where \( m \) is the mass of the object and \( g \) is the gravitational acceleration (approximately \( 9.8 \, \text{m/s}^2 \)).
In the case of our elevator, which has a mass of \( 1200 \, \text{kg} \), the weight or gravitational force acting on it is \( 11760 \, \text{N} \).
This weight is significant because it is the force that the cable's tension must overcome to lift the elevator upwards. Keeping the force equal to the weight ensures that the elevator moves at a constant velocity. Understanding the elevator's weight is crucial in calculating the work done to lift the elevator, especially when the motion is constant and no additional acceleration force is involved.
Constant Velocity Motion
Constant velocity motion is a state where an object moves in a straight line at a consistent speed. For an elevator being lifted at a constant velocity, this means that the forces are perfectly balanced. The upward force (tension in the cable) is equal to the downward force (gravitational force of the elevator's weight).
This concept is vital because it implies that no net force is acting on the elevator.
  • The tension force equals the gravitational force, ensuring no acceleration occurs.
  • The net work done is zero, as no additional energy is needed to change the elevator's speed.
What makes this situation interesting and—often encountered in physics problems—is how the work done by opposing forces (like tension and gravity) becomes equal in magnitude and opposite in direction. Even though work is being continuously done to maintain motion, constant velocity signifies a delicate balance where input and resistance forces are evenly matched.

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Most popular questions from this chapter

Juggles and Bangles are clowns. Juggles stands on one end of a teeter-totter at rest on the ground. Bangles jumps off a platform \(2.5 \mathrm{m}\) above the ground and lands on the other end of the teeter-totter, launching Juggles into the air. Juggles rises to a height of \(3.3 \mathrm{m}\) above the ground, at which point he has the same amount of gravitational potential energy as Bangles had before he jumped, assuming both potential energies are measured using the ground as the reference level. Bangles' mass is \(86 \mathrm{kg}\). What is Juggles' mass?

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The concepts in this problem are similar to those in Multiple-Concept Example 4, except that the force doing the work in this problem is the tension in the cable. A rescue helicopter lifts a \(79-\mathrm{kg}\) person straight up by means of a cable. The person has an upward acceleration of \(0.70 \mathrm{m} / \mathrm{s}^{2}\) and is lifted from rest through a distance of \(11 \mathrm{m}\). (a) What is the tension in the cable? How much work is done by (b) the tension in the cable and (c) the person's weight? (d) Use the work-energy theorem and find the final speed of the person.

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