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Chapter 7: Problem 2

A tennis player hits a \(58.0 \mathrm{~g}\) tennis ball so that it goes straight up and reaches a maximum height of \(6.17 \mathrm{~m} .\) How much work does gravity do on the ball on the way up? On the way down?

Short Answer

Expert verified
Gravity does -3.51 J on the way up and +3.51 J on the way down.

Step by step solution

01

Identify Given Values

The mass of the tennis ball is given as \( m = 58.0 \, \text{g} = 0.058 \, \text{kg} \) and it reaches a maximum height of \( h = 6.17 \, \text{m} \).
02

Understand the Work-Energy Principle

The work done by gravity on the ball is equal to the change in gravitational potential energy when the ball is lifted to a height h, which is calculated using the formula \( W = mgh \), where \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity.
03

Calculate Work Done by Gravity on the Way Up

To find the work done by gravity on the way up, use the formula: \[ W_{up} = -mgh = -(0.058 \, \text{kg})(9.81 \, \text{m/s}^2)(6.17 \, \text{m}) \] Solving gives \( W_{up} = -3.51 \, \text{J} \).
04

Calculate Work Done by Gravity on the Way Down

On the way down, the work done by gravity is equal to the magnitude of the work on the way up, but positive. Thus, \[ W_{down} = +3.51 \, \text{J} \].

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Key Concepts

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

Physics Problems
Physics problems can often seem complex, but they are a great way to understand physical laws more deeply. They help to practically apply theories and formulas to real-world situations. When approaching a physics problem, it is important to:
  • Identify the known values such as mass, volume, or height.
  • Understand which physical laws apply, such as Newton's Laws or the Conservation of Energy.
  • Use relevant formulas to find unknown quantities by plugging in the known values.
  • Pay attention to units and convert them appropriately, ensuring they are consistent when doing calculations.
In the case of our tennis ball problem, identifying the mass and height was crucial, as well as applying the understanding of how gravitational force works on objects in motion. Solving such problems hones critical thinking and problem-solving skills, which are essential not just in physics, but in many areas of life.
Work-Energy Principle
The work-energy principle is a fundamental concept in physics that relates work done on an object to its change in energy. This principle states that the work done by all forces acting on a particle is equal to the change in its kinetic energy. In simpler terms, it means:
  • When work is done on an object, it causes changes in energy, such as converting potential energy to kinetic energy.
  • The formula to calculate work done is given by \( W = F imes d \times \cos(\theta) \), where \( W \) is work, \( F \) is force, \( d \) is displacement, and \( \theta \) is the angle between the force and displacement vectors.
  • Gravitational work, a specific case of the work-energy principle, involves changes in gravitational potential energy.
In our tennis ball scenario, the force exerted by gravity, the height the ball reaches, and the angle (which in this case is directly vertical) are all considered to calculate work done. This principle aids in understanding energy transfer when objects are in motion or when forces like gravity are applied.
Gravitational Potential Energy
Gravitational potential energy (GPE) is a type of energy an object possesses due to its position in a gravitational field. It depends on three key factors:
  • The mass of the object (\( m \)), measured in kilograms (kg).
  • The height (\( h \)) above a reference point, usually measured in meters (m).
  • The gravitational acceleration (\( g \)), typically approximated as \( 9.81 \, \text{m/s}^2 \) on Earth.
The formula to calculate gravitational potential energy is:\[ GPE = mgh \]This energy represents the potential for an object to do work due to its elevated position. As demonstrated in the tennis ball problem, the ball's gravitational potential energy increases as it rises, requiring work by gravity when it moves up. Conversely, as it falls, this potential energy is converted back into kinetic energy, showcasing the dynamic interplay between different forms of energy due to gravity's influence.

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Most popular questions from this chapter

A \(250 \mathrm{~g}\) object on a frictionless, horizontal lab table is pushed against a spring of force constant \(35 \mathrm{~N} / \mathrm{cm}\) and then released. Just before the object is released, the spring is compressed \(12.0 \mathrm{~cm} .\) How fast is the object moving when it has gained half of the spring's original stored energy?

A ball is thrown upward with an initial velocity of \(15 \mathrm{~m} / \mathrm{s}\) at an angle of \(60.0^{\circ}\) above the horizontal. Use energy conservation to find the ball's greatest height above the ground.

Your friend (mass \(65.0 \mathrm{~kg}\) ) is standing on the ice in the middle of a frozen pond. There is very little friction between her feet and the ice, so she is unable to walk. Fortunately, a light rope is tied around her waist and you stand on the bank holding the other end. You pull on the rope for \(3.00 \mathrm{~s}\) and accelerate your friend from rest to a speed of \(6.00 \mathrm{~m} / \mathrm{s}\) while you remain at rest. What is the average power supplied by the force you applied?

On flat ground, a \(70 \mathrm{~kg}\) person requires about \(300 \mathrm{~W}\) of metabolic power to walk at a steady pace of \(5.0 \mathrm{~km} / \mathrm{h}(1.4 \mathrm{~m} / \mathrm{s})\) Using the same metabolic power output, that person can bicycle over the same ground at \(15 \mathrm{~km} / \mathrm{h}\). A \(70 \mathrm{~kg}\) person walks at a steady pace of \(5.0 \mathrm{~km} / \mathrm{h}\) on a treadmill at a \(5.0 \%\) grade (that is, the vertical distance covered is \(5.0 \%\) of the horizontal distance covered). If we assume the metabolic power required is equal to that required for walking on a flat surface plus the rate of doing work for the vertical climb, how much power is required? A. \(300 \mathrm{~W}\) B. 315 W C. \(350 \mathrm{~W}\) D. \(370 \mathrm{~W}\)

A rope is tied to a box and used to pull the box \(1.5 \mathrm{~m}\) along a h zontal floor. The rope makes an angle of \(30^{\circ}\) with the horizontal has a tension of \(5 \mathrm{~N}\). The opposing friction force between the box the floor is \(1 \mathrm{~N}\). How much work does each of the following forces do on the box: (a) gravity, (b) the tension in the rope, (c) friction, and (d) the normal force? What is the total work done on the box?
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