Problem 24 - Thomas Calculus

Question

Tennis \(A\) -oz tennis ball was served at 160 \(\mathrm{ft} / \mathrm{sec}\) (about 109 \(\mathrm{mph}\) ). How much work was done on the ball to make it go this fast? (To find the ball's mass from its weight, express the weight in pounds and divide by \(32 \mathrm{ft} / \mathrm{sec}^{2},\) the acceleration of gravity.)

Step-by-Step Solution

Answer

The work done on the ball is \(25A\) ft-lbs.

Step 1: Convert ounces to pounds

First, we convert the weight of the tennis ball from ounces to pounds. Since there are 16 ounces in one pound, if the tennis ball weighs \( A \) ounces, it weighs \( A/16 \) pounds.

Step 2: Calculate the mass of the tennis ball

To find the mass of the ball, divide the weight in pounds by the acceleration due to gravity, which is \( 32 \, \mathrm{ft/s}^2 \). Therefore, the mass \( m \) of the ball is \( m = \frac{A}{16 \times 32} = \frac{A}{512} \) slugs.

Step 3: Calculate the kinetic energy

The work done on the ball is equal to its kinetic energy when it is moving at the given speed. The formula for kinetic energy is \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass and \( v \) is the velocity in \( \mathrm{ft/s} \). Thus, the kinetic energy is \( KE = \frac{1}{2} \times \frac{A}{512} \times 160^2 \).

Step 4: Final computation of work done

Substitute the values into the kinetic energy formula: \( KE = \frac{1}{2} \times \frac{A}{512} \times (160)^2 = \frac{A \times 25600}{1024} = \frac{25A}{1} = 25A \). Thus, the work done on the ball is \( 25A \) ft-lbs.

Key Concepts

Understanding Units Conversion

In problem-solving, converting units is a key step, especially when dealing with real-world measurements. Let's say we have a tennis ball weighing in ounces, and we need to perform calculations that require the weight in pounds. Here's how you do it. Since 1 pound equals 16 ounces, you convert by dividing the number of ounces by 16. For instance, if the ball weighs \( A \) ounces, the weight in pounds is \( A/16 \).
This step is essential to ensure that your calculations stay consistent and accurate. Always check that the units match up with what's expected in your formulas.

• 1 pound = 16 ounces
• Weight in pounds = Weight in ounces / 16

By understanding and applying units conversion correctly, you're setting a strong foundation for the entire problem-solving process.

Exploring Kinetic Energy

Kinetic energy is a measure of the energy that an object has due to its motion. It can be expressed mathematically as \( KE = \frac{1}{2} m v^2 \), where \( m \) is the mass of the object and \( v \) is its velocity. The unit of kinetic energy is foot-pounds (ft-lbs) in this context, which is often used in physics to describe energy.
In the case of the tennis ball, once we know its mass and velocity, we can calculate the kinetic energy to determine how much work has been done to get the ball moving at a certain speed. This involves substituting the known values into the kinetic energy formula.

• \( KE = \frac{1}{2} m v^2 \)
• Mass (\( m \)) should be in slugs
• Velocity (\( v \)) should be in ft/s

Understanding kinetic energy helps you grasp how motion and mass interplay to create energy in moving objects.

Demystifying Mass Calculation

To find the mass in this exercise, you must first understand the relationship between weight and mass. Weight is force, typically measured in pounds, and depends on the acceleration due to gravity. Mass, on the other hand, is a measure of the amount of matter in an object and is measured in slugs when using the imperial system.
The mass of an object is calculated by dividing its weight (in pounds) by the acceleration due to gravity, which is \( 32\, \mathrm{ft/s}^2 \). This gives us mass in slugs. In mathematical terms, if a ball's weight in pounds is \( W \), its mass \( m \) is \( m = \frac{W}{32} \).

• Weight to mass conversion: divide by \( 32\, \mathrm{ft/s}^2 \)
• Mass unit: slugs
• Importance of consistency in units

By consistently applying this method, you ensure your calculations remain accurate and align with how mechanical systems work in physics.