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Chapter 7: Problem 46

The loaded cab of an elevator has a mass of \(3.0 \times 10^{3} \mathrm{~kg}\) and moves \(210 \mathrm{~m}\) up the shaft in \(23 \mathrm{~s}\) at constant speed. At what average rate does the force from the cable do work on the cab?

Short Answer

Expert verified
The average rate is approximately 2.691 MW.

Step by step solution

01

Understand the Problem

We need to find the average rate at which the force from the cable does work on the elevator cab. The cab moves at a constant speed, which means the net force is zero due to balanced forces (gravity and cable force). The work done on the elevator is the force exerted by the cable over the distance it travels.
02

Calculate Gravitational Force

The gravitational force (weight) acting on the cab can be calculated using the formula: \[ F = m imes g \] where \( m = 3.0 \times 10^3 \, \text{kg} \) is the mass of the cab and \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity.
03

Set Up the Work Equation

The work done by the cable can be calculated using the formula: \[ W = F imes d \] where \( F \) is the force exerted by the cable (equal to the gravitational force to maintain constant speed) and \( d = 210 \, \text{m} \) is the distance the cab travels.
04

Calculate the Work Done

Substitute the force and distance into the work formula: \[ W = (3.0 \times 10^3 \, \text{kg} \times 9.8 \, \text{m/s}^2) \times 210 \, \text{m} \]Compute the result to find the total work done by the cable.
05

Calculate the Average Power

The average power, or the rate of work done, can be calculated using the formula:\[ P = \frac{W}{t} \] where \( t = 23 \, \text{s} \) is the time period over which the work is done.
06

Compute the Average Power

Substitute the total work from Step 4 and the time into the power formula:\[ P = \frac{(3.0 \times 10^3 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 210 \, \text{m})}{23 \, \text{s}} \]Calculate the result to find the average power exerted by the cable.

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Key Concepts

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

Gravitational Force
In physics, gravitational force is the attractive force between any two masses. It's one of the four fundamental forces in nature. For this specific exercise, the focus is on the gravitational force exerted on the elevator cab. The formula to calculate gravitational force is simple:
  • The force, \( F \), exerted by gravity is equal to the product of mass \( m \) and the acceleration due to gravity \( g \).
  • Mathematically, this is expressed as \( F = m \times g \).
In this problem:
  • The mass of the elevator cab \( m \) is \( 3.0 \times 10^3 \text{ kg} \).
  • The acceleration due to gravity \( g \) is \( 9.8 \text{ m/s}^2 \).
By substituting these values into the equation, one can determine the force acting downward due to gravity. This force is critical as it counterbalances the force from the cable to keep the cab moving at constant speed. Without gravity's pull, the work done calculations would be different.
Average Power
Average power is a measure of the rate at which work is done or energy is transferred. It's relevant in situations where a force is applied over a period of time, like in this elevator scenario. Here's how you can compute average power:
  • Average power, \( P \), is calculated as the total work \( W \) done divided by the time \( t \) it takes to do that work.
  • The formula is \( P = \frac{W}{t} \).
In the case of the elevator:
  • The work done by the cable force is calculated first, as seen in the gravitational force section.
  • And, the given time is \( 23 \text{ s} \).
After you've found the total work done (taking into account gravity and distance), dividing by the time will yield the average power. Remember, power tells us how fast energy is being used or converted, and in this scenario, how efficiently the elevator's cable moves the cab over a period.
Constant Speed Motion
When an object moves at constant speed, it means that the net force acting on the object is zero. This concept comes into play prominently in understanding the work done by the elevator cable in our exercise. Here is the rationale behind constant speed motion:
  • If an object like an elevator cab moves at constant speed, it means the forces on it are balanced.
  • In this scenario, the gravitational force pulling down on the cab is exactly equal to the force exerted by the cable pulling it upwards.
  • Therefore, the cab does not accelerate. It maintains the same speed throughout its journey.
The significance of constant speed is that the work done by the cable is exactly equal to the gravitational force times the distance moved, since there's no additional work done to increase speed or overcome net acceleration. This makes the problem simpler to solve, focusing solely on balancing the gravitational force.

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Most popular questions from this chapter

A horse pulls a cart with a force of \(40 \mathrm{lb}\) at an angle of \(30^{\circ}\) above the horizontal and moves along at a speed of \(6.0 \mathrm{mi} / \mathrm{h} .\) (a) How much work does the force do in \(10 \mathrm{~min} ?\) (b) What is the average power (in horsepower) of the force?

On August 10,1972, a large meteorite skipped across the atmosphere above the western United States and western Canada, much like a stone skipped across water. The accompanying fireball was so bright that it could be seen in the daytime sky and was brighter than the usual meteorite trail. The meteorite's mass was about \(4 \times 10^{6} \mathrm{~kg}\); its speed was about \(15 \mathrm{~km} / \mathrm{s}\). Had it entered the atmosphere vertically, it would have hit Earth's surface with about the same speed. (a) Calculate the meteorite's loss of kinetic energy (in joules) that would have been associated with the vertical impact. (b) Express the energy as a multiple of the explosive energy of 1 megaton of TNT, which is \(4.2 \times 10^{15} \mathrm{~J}\). (c) The energy associated with the atomic bomb explosion over Hiroshima was equivalent to 13 kilotons of TNT. To how many Hiroshima bombs would the meteorite impact have been equivalent?

A cord is used to vertically lower an initially stationary block of mass \(M\) at a constant downward acceleration of \(g / 4\). When the block has fallen a distance \(d\), find (a) the work done by the cord's force on the block, (b) the work done by the gravitational force on the block, (c) the kinetic energy of the block, and (d) the speed of the block.

A spring with a spring constant of \(18.0 \mathrm{~N} / \mathrm{cm}\) has a cage attached to its free end. (a) How much work does the spring force do on the cage when the spring is stretched from its relaxed length by \(7.60 \mathrm{~mm} ?\) (b) How much additional work is done by the spring force when the spring is stretched by an additional \(7.60 \mathrm{~mm} ?\)

A particle moves along a straight path through displacement \(\vec{d}=(8 \mathrm{~m}) \hat{\mathrm{i}}+c \hat{\mathrm{j}}\) while force \(\vec{F}=(2 \mathrm{~N}) \hat{\mathrm{i}}-(4 \mathrm{~N}) \hat{\mathrm{j}}\) acts on it. (Other forces also act on the particle.) What is the value of \(c\) if the work done by \(\vec{F}\) on the particle is (a) zero, (b) positive, and (c) negative?
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