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A cable lifts a \(1200-\mathrm{kg}\) 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
(a) Work done by tension: 411600 J; (b) Work done by weight: -411600 J.

Step by step solution

01

Understand Work Done

Work is defined as the force applied over a distance. The formula to calculate work done, when force is applied in the direction of displacement, is given by \( W = F \cdot d \), where \( F \) is the force and \( d \) is the distance over which the force is applied.
02

Identify Forces Involved

There are two main forces acting on the elevator: - The tension in the cable that lifts the elevator upwards. - The gravitational force (weight) pulling the elevator down. Since these forces act in opposite directions, we will consider each separately.
03

Calculate Tension Force

The elevator moves at a constant velocity, which means the net force is zero. Therefore, the tension in the cable (\( T \)) equals the gravitational force on the elevator. The gravitational force (weight) is calculated as \( F_g = mg \), where \( m = 1200 \) kg and \( g = 9.8 \, \mathrm{m/s^2} \). \[ F_g = 1200 \, \mathrm{kg} \times 9.8 \, \mathrm{m/s^2} = 11760 \, \mathrm{N} \].Thus, the tension \( T = 11760 \, \mathrm{N} \).
04

Calculate Work Done by Tension

The work done by the tension is given by \( W_{\text{tension}} = T \times d \), where \( d = 35 \, \mathrm{m} \). \[ W_{\text{tension}} = 11760 \, \mathrm{N} \times 35 \, \mathrm{m} = 411600 \, \mathrm{J} \].
05

Calculate Work Done by the Elevator's Weight

The work done by the gravitational force is calculated using \( W_{\text{gravity}} = - F_g \times d \), since the force acts in the opposite direction of the displacement. \[ W_{\text{gravity}} = -11760 \, \mathrm{N} \times 35 \, \mathrm{m} = -411600 \, \mathrm{J} \].

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Key Concepts

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

Tension Force
Tension force is a common phenomenon in physics, encountered in situations where a cable, rope, or similar object is used to lift or hold an object. In the context of an elevator, the tension force is the force exerted by the cable on the elevator to lift it upward. When considering tension, it's important to understand that this force acts along the length of the cable in a direction that attempts to pull apart the ends of the cable.

In our scenario, the elevator is being lifted at a constant velocity. This means the tension force is perfectly balancing the gravitational force. To achieve this, the magnitude of the tension force matches the force of gravity acting on the elevator. This results in a net force of zero, allowing the elevator to smoothly operate without acceleration or deceleration.

Key points about tension force include:
  • Acts along the length of the cable.
  • Is equal and opposite to the force of gravity when moving at constant velocity.
  • Calculated easily using the weight (gravitational force) of the object being lifted.
Gravitational Force
Gravitational force plays a crucial role in elevator physics, just as it does in numerous other physics problems. It is the force with which the Earth pulls objects towards its center, which gives them weight. This force is inherently directed downward towards the Earth's surface.

For our elevator problem, calculating gravitational force involves using the equation: \[ F_g = m \cdot g \]where:
  • \( m \) is the mass of the object (elevator) in kilograms.
  • \( g \) is the acceleration due to gravity, approximately \( 9.8 \mathrm{m/s^2} \).
Subsequently, the gravitational force acting on the 1200 kg elevator is \( 11760 \mathrm{N} \).

This gravitational force opposes the motion of the elevator as it ascends, meaning any additional force, such as tension, must counteract it to allow for constant velocity movement.
Constant Velocity Motion
When an object moves with constant velocity, it implies that its speed and direction remain unchanged over time. This occurs when the net force acting on the object is zero, ensuring that there is a perfect balance of forces.

In our elevator scenario, moving with constant velocity means:
  • The upward tension force equals the downward gravitational force.
  • There is no acceleration or deceleration of the elevator.
  • Work done by both forces gets counteracted as they act in opposite directions, resulting in zero net work affecting the elevator's velocity.
This balanced state is significant since it helps ensure smooth and efficient operation of lifts, preventing unnecessary force and energy expenditure.
Elevator Physics
Elevator physics combines multiple fundamental concepts in mechanics, including tension force, gravitational force, and the principles of motion. An elevator system's design aims to ensure safe and efficient movement of people or goods over specific vertical distances.

The operation of an elevator involves:
  • Synchronization between tension and gravitational forces to maintain constant velocity.
  • Calculations of work done to determine the energy required for motion.
  • Safety mechanisms to manage force and ensure steady movement.
Understanding these principles in elevator physics allows for reliable and predictable elevator operation, which is critical in multi-story buildings. The ideal scenario is one where forces like tension and gravity are managed effectively, ensuring that the elevator meets intended speeds and stops accurately at each required floor.

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Most popular questions from this chapter

You are trying to lose weight by working out on a rowing machine. Each time you pull the rowing bar (which simulates the "oars") toward you, it moves a distance of \(1.2 \mathrm{~m}\) in a time of \(1.5 \mathrm{~s}\). The readout on the display indicates that the average power you are producing is \(82 \mathrm{~W}\). What is the magnitude of the force that you exert on the handle?

A basketball of mass \(0.60 \mathrm{~kg}\) is dropped from rest from a height of \(1.05 \mathrm{~m}\). It rebounds to a height of \(0.57 \mathrm{~m}\) (a) How much mechanical energy was lost during the collision with the floor? (b) A basketball player dribbles the ball from a height of \(1.05 \mathrm{~m}\) by exerting a constant downward force on it for a distance of \(0.080 \mathrm{~m}\). In dribbling, the player compensates for the mechanical energy lost during each bounce. If the ball now returns to a height of 1.05 \(\mathrm{m},\) what is the magnitude of the force?

A \(3.00-\mathrm{kg}\) model rocket is launched straight up. It reaches a maximum height of \(1.00 \times 10^{2} \mathrm{~m}\) above where its engine cuts out, even though air resistance performs \(-8.00 \times 10^{2} \mathrm{~J}\) of work on the rocket. What would have been this height if there were no air resistance?

Multiple-Concept Example 13 presents useful background for this problem. The cheetah is one of the fastest-accelerating animals, because it can go from rest to \(27 \mathrm{~m} / \mathrm{s}\) (about 60 \(\mathrm{mi} / \mathrm{h}\) ) in \(4.0 \mathrm{~s}\). If its mass is \(110 \mathrm{~kg}\), determine the average power developed by the cheetah during the acceleration phase of its motion. Express your answer in (a) watts and (b) horsepower.

A pitcher throws a \(0.140-\mathrm{kg}\) baseball, and it approaches the bat at a speed of \(40.0 \mathrm{~m} / \mathrm{s}\). The bat does \(W_{n c}=70.0 \mathrm{~J}\) of work on the ball in hitting it. Ignoring air resistance, determine the speed of the ball after the ball leaves the bat and is \(25.0 \mathrm{~m}\) above the point of impact.

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