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An automobile is pushed \(10.0 \mathrm{ft}\) by a woman exerting \(80.0 \mathrm{lb}\) of force horizontally on the vehicle. How much work does she do (a) in \(\mathrm{ft} \cdot \mathrm{lb}\) and \((b)\) in joules?

Short Answer

Expert verified
The work done is 800.0 ft·lb and approximately 1084.64 joules.

Step by step solution

01

Understanding the Problem

To find the work done by the woman, we need to apply the formula for work:\[ W = F imes d \]where \( W \) is the work done, \( F \) is the force applied, and \( d \) is the distance over which the force is applied. Here, \( F = 80.0 \) lb and \( d = 10.0 \) ft.
02

Converting Units

Since part b of the problem requires the work to be in joules, we need to convert from \( \mathrm{ft} \cdot \mathrm{lb} \) to joules. Recall the conversion factor: 1 foot-pound is equivalent to approximately 1.3558 joules.
03

Calculate Work in Foot-Pounds

Using the formula, calculate the work done in foot-pounds:\[ W = 80.0 \, \text{lb} \times 10.0 \, \text{ft} = 800.0 \, \text{ft}\cdot\text{lb} \]This is the work done in part (a).
04

Convert Work to Joules

To convert the work from foot-pounds to joules, multiply by the conversion factor:\[ W = 800.0 \, \text{ft}\cdot\text{lb} \times 1.3558 \, \frac{\text{joules}}{\text{ft}\cdot\text{lb}} \approx 1084.64 \, \text{joules} \]This is the work done in part (b).

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Key Concepts

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

Force and Motion
In physics, force is any interaction that, when unopposed, will change the motion of an object. Force can cause an object at rest to move, or it can accelerate a moving object by changing its speed or direction. Motion is the change in position of an object with respect to time. It describes the path that an object takes as well as how fast it is going.

When solving problems about force and motion, like the example of pushing an automobile, it's essential to know how force and distance relate to work. Work done on an object is calculated as the product of the force applied to it and the distance over which that force is applied.
  • Force is often measured in pounds or newtons.
  • Motion is measured in terms of distance, such as feet or meters.
Here, the woman exerts a force of 80 lb over a distance of 10 ft horizontally. Knowing how force causes motion provides a basis for calculating the work done, which is simply the product of these two quantities: 80 lb of force applied over a 10 ft distance equals 800 ft·lb of work.
Unit Conversion
Unit conversion is an essential skill in physics as it allows you to interpret measurements across different unit systems. The International System of Units (SI) is the standard metric system used globally, where force is measured in newtons and work in joules. However, in places like the United States, the imperial system is commonly used for certain applications.

To solve physics problems, like calculating work in joules from foot-pounds, you need to understand the conversion factors. Here are the basics for unit conversion in this context:
  • 1 foot-pound (ft·lb) is approximately equal to 1.3558 joules (J).
  • Conversely, 1 joule is roughly 0.7376 foot-pounds.
So, after calculating the work in foot-pounds, as was done in the exercise, you multiply by this conversion factor to find the joules. In our example, 800 ft·lb converts to approximately 1084.64 joules, showcasing the power of correct unit conversion to accurately relay information between systems.
Physics Problems
Physics problems can seem daunting, but breaking them into steps makes them manageable. The key is understanding and applying the underlying principles. Take work and energy problems as an illustration. Begin by identifying what you need to find—such as the amount of work done in different units as we did here.

The approach involves:
  • Identifying the physical principles that apply (e.g., work-energy principle).
  • Writing down known values and what needs to be calculated.
  • Applying the right formulas and conversion factors.
This framework allows you to solve complex problems by understanding the physics involved. For example, knowing how the force exerted relates directly to the distance moved helps determine the work done. Converting results into different units, like joules, ensures that results meet specified requirements. Whether you’re solving for foot-pounds or joules, understanding these basic principles will guide you through the solution process.

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Most popular questions from this chapter

A pump lifts water from a lake to a large tank \(20 \mathrm{~m}\) above the lake. How much work against gravity does the pump do as it transfers \(5.0 \mathrm{~m}^{3}\) of water to the tank? One cubic meter of water has a mass of \(1000 \mathrm{~kg}\).

An advertisement claims that a certain 1200 -kg car can accelerate from rest to a speed of \(25 \mathrm{~m} / \mathrm{s}\) in a time of \(8.0 \mathrm{~s}\). What average power must the motor develop to produce this acceleration? Give your answer in both watts and horsepower. Ignore friction losses. The work done in accelerating the car is $$ \text { Work done }=\text { Change in } \mathrm{KE}=\frac{1}{2} m\left(v_{f}^{2}-v_{i}^{2}\right)=\frac{1}{2} m v_{f}^{2} $$ The time taken for this work to be performed is \(8.0 \mathrm{~s}\). Therefore, to two significant figures, $$ \text { Power }=\frac{\text { Work }}{\text { Time }}=\frac{\frac{1}{2}(1200 \mathrm{~kg})(25 \mathrm{~m} / \mathrm{s})^{2}}{8.0 \mathrm{~s}}=46875 \mathrm{~W}=47 \mathrm{~kW} $$ Converting from watts to horsepower, we have $$ \text { Power }=(46875 \mathrm{~W})\left(\frac{1 \mathrm{hp}}{746 \mathrm{~W}}\right)=63 \mathrm{hp} $$

A 1200-kg car going \(30 \mathrm{~m} / \mathrm{s}\) applies its brakes and skids to rest. If the friction force between the sliding tires and the pavement is \(6000 \mathrm{~N}\), how far does the car skid before coming to rest?

A force of \(3.0 \mathrm{~N}\) acts through a distance of \(12 \mathrm{~m}\) in the direction of the force. Find the work done.

A 2000 -kg elevator rises from rest in the basement to the fourth floor, a distance of \(25 \mathrm{~m}\). As it passes the fourth floor, its speed is \(3.0 \mathrm{~m} / \mathrm{s}\). There is a constant frictional force of \(500 \mathrm{~N}\). Calculate the work done by the lifting mechanism.

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