Problem 2
Question
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?
Step-by-Step Solution
Verified Answer
-3.50 J on the way up; 3.50 J on the way down.
1Step 1: 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.
2Step 2: 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 \( m = \frac{58.0}{1000} \text{ kg} = 0.058 \text{ kg} \).
3Step 3: 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.
4Step 4: 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:\[ W_{up} = -mgh = -(0.058 \text{ kg})(9.81 \text{ m/s}^2)(6.17 \text{ m}) \]\[ W_{up} = -3.50 \text{ J} \].
5Step 5: 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:\[ W_{down} = mgh = 3.50 \text{ J} \].
6Step 6: 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.
Key Concepts
Gravitational Potential EnergyMass ConversionNegative Work by GravityPositive Work by Gravity
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.
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.
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/s²),
- \( h \) is the height (6.17 m).
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:
- \( m = 0.058 \, \text{kg} \),
- \( g = 9.81 \, \text{m/s}^2 \),
- \( h = 6.17 \, \text{m} \).
Other exercises in this chapter
Problem 1
\(\cdot\) A fisherman reels in 12.0 \(\mathrm{m}\) of line while landing a fish, using a constant forward pull of 25.0 \(\mathrm{N}\) . How much work does the t
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\(\cdot\) A constant horizontal pull of 8.50 \(\mathrm{N}\) drags a box along a hor- izontal floor through a distance of 17.4 \(\mathrm{m} .\) (a) How much work
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\(\cdot\) You push your physics book 1.50 \(\mathrm{m}\) along a horizontal tabletop with a horizontal push of 2.40 \(\mathrm{N}\) while the opposing force of f
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