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Chapter 7: Problem 2

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

Short Answer

Expert verified
-3.50 J on the way up; 3.50 J on the way down.

Step by step solution

01

Understanding the Problem

We are asked to calculate the work done by gravity on a tennis ball of mass 58.0 g as it travels up to a height of 6.17 m and back down to its initial position. The work done by gravity depends on the change in gravitational potential energy.
02

Convert Mass into Kilograms

First, convert the mass of the tennis ball from grams to kilograms since the standard unit of mass in physics calculations is kilograms. This conversion is given by \( 1 \text{ kg} = 1000 \text{ g} \). Thus, the mass \( m = \frac{58.0}{1000} \text{ kg} = 0.058 \text{ kg} \).
03

Use the Work-Energy Principle

The work done by gravity is equal to the change in gravitational potential energy. The formula to find gravitational potential energy is \[ U = mgh \]where \( m \) is the mass, \( g \) is the acceleration due to gravity \( (9.81 \text{ m/s}^2) \), and \( h \) is the height.
04

Calculate Work Done On the Way Up

On the way up, gravity does negative work on the ball as the ball gains potential energy. The potential energy change can be calculated using the initial value of height as 0 m and the final value as 6.17 m:\[ W_{up} = -mgh = -(0.058 \text{ kg})(9.81 \text{ m/s}^2)(6.17 \text{ m}) \]\[ W_{up} = -3.50 \text{ J} \].
05

Calculate Work Done On the Way Down

On the way down, gravity does positive work, equal but opposite in sign to the work done on the way up because the ball returns to its initial height. Thus, the work done on the way down is:\[ W_{down} = mgh = 3.50 \text{ J} \].
06

Conclusion

The amount of work gravity does on the ball is \(-3.50 \text{ J}\) on the way up and \(3.50 \text{ J}\) on the way down.

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Key Concepts

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

Gravitational Potential Energy
Gravitational potential energy is the energy held by an object because of its position relative to a lower point, typically Earth. It is an essential concept when dealing with objects moving in a vertical direction. This energy is calculated using the formula \( U = mgh \), where:
  • \( m \) is the mass of the object,
  • \( g \) is the acceleration due to gravity, approximately \( 9.81 \, \text{m/s}^2 \),
  • \( h \) is the height above the reference point.
This formula calculates the energy due to an object's position. When the height of an object increases, so does its gravitational potential energy, requiring work to lift the object. Conversely, when the height decreases, energy is released. This principle is common in various physics problems involving vertical motion of objects, like our tennis ball example.
Mass Conversion
In physics calculations, it's crucial to work with standard units. For mass, the standard unit is kilograms, not grams. This means any mass expressed in grams must first be converted. To convert grams to kilograms, divide by 1000. In the case of the 58.0 g tennis ball, \( 58.0 \, \text{g} = 0.058 \, \text{kg} \).
Using kilograms ensures uniformity across equations, which often involve standard units like meters for distance and seconds for time. This conversion helps avoid errors and aligns with international standards in scientific calculations, providing clarity and precision, as shown in many physics exercises.
Negative Work by Gravity
When a force, such as gravity, acts opposite to the direction of an object's movement, it is performing negative work. This happens when you lift an object upward, against gravity. In our example with the tennis ball, as it goes up, gravity works against it. This is calculated using \( W = -mgh \).
  • \( m \) is the ball's mass (0.058 kg),
  • \( g \) is the acceleration due to gravity (9.81 m/s虏),
  • \( h \) is the height (6.17 m).
Therefore, the negative work done by gravity is \( -3.50 \, \text{J} \). It shows that energy is required to perform the work needed to raise the ball, stored as gravitational potential energy.
Positive Work by Gravity
Positive work occurs when a force acts in the direction of an object's movement. As the tennis ball falls back to its starting point, gravity assists in its motion. Thus, gravity performs positive work. The calculation remains \( W = mgh \), but it represents energy being released:
  • \( m = 0.058 \, \text{kg} \),
  • \( g = 9.81 \, \text{m/s}^2 \),
  • \( h = 6.17 \, \text{m} \).
This results in \( 3.50 \, \text{J} \) of work done by gravity. During the descent, the gravitational potential energy converts back to kinetic energy. This type of positive work confirms gravity's role in conservation of energy principles, illustrating how energy transitions in free-falling motion.

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Most popular questions from this chapter

\(\bullet\) How high can we jump? The maximum height a typical human can jump from a crouched start is about 60 \(\mathrm{cm} .\) By how much does the gravitational potential energy increase for a 72 \(\mathrm{kg}\) person in such a jump? Where does this energy come from?

\(\bullet\) \(\bullet\) \(\bullet\) Pendulum. A small 0.12 kg metal ball is tied to a very light (essentially massless) string 0.80 \(\mathrm{m}\) long to form a pendu- lum that is then set swinging by releasing the ball from rest when the string makes a \(45^{\circ}\) angle with the vertical. Air drag and other forms of friction are negligible. What is the speed of the ball when the string passes through its vertical position and what is the tension in the string at that instant?

\(\bullet\) Tarzan and Jane. Tarzan, in one tree, sights Jane in another tree. He grabs the end of a vine with length 20 \(\mathrm{m}\) that makes an angle of \(45^{\circ}\) with the vertical, steps off his tree limb, and swings down and then up to Jane's open arms. When he arrives, his vine makes an angle of \(30^{\circ}\) with the vertical. Deter- mine whether he gives her a tender embrace or knocks her off her limb by calculating Tarzan's speed just before he reaches Jane. You can ignore air resistance and the mass of the vine.

\(\cdot\) \(\cdot\) You throw a 20 \(\mathrm{N}\) rock into the air from ground level and observe that, when it is 15.0 \(\mathrm{m}\) high, it is traveling upward at 25.0 \(\mathrm{m} / \mathrm{s} .\) Use the work-energy principle to find (a) the rock's speed just as it left the ground and (b) the maximum height the rock will reach.

\(\bullet\) \(\bullet\) Energy requirements of the body. A 70 \(\mathrm{kg}\) human uses energy at the rate of \(80 \mathrm{J} / \mathrm{s},\) on average, for just resting and sleeping. When the person is engaged in more strenuous activities, the rate can be much higher. (a) If the individual did nothing but rest, how many food calories per day would she or he have to eat to make up for those used up? (b) In what forms is energy used when a person is resting or sleep- ing? In other words, what happens to those 80 \(\mathrm{J} / \mathrm{s} ?\) Hint: What kinds of energy, mechanical and otherwise, do our body components have?) (c) If an average person rested and did other low-level activity for 16 hours (which consumes 80 \(\mathrm{J} / \mathrm{s} )\) and did light activity on the job for 8 hours (which consumes \(200 \mathrm{J} / \mathrm{s} ),\) how many calories would she or he have to con- sume per day to make up for the energy used up?
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