Problem 26 - Chemistry: The Central Science in SI Units, Global Edition

Question

(a) A baseball weighs \(145.4 \mathrm{~g}\). What is the kinetic energy, in joules, of this baseball when it is thrown by a major league pitcher at \(150 \mathrm{~km} / \mathrm{h} ?\) (b) By what factor will the kinetic energy change if the speed of the baseball is decreased to \(90 \mathrm{~km} / \mathrm{h} ?\) (c) What happens to the kinetic energy when the baseball is caught by the catcher? Is it converted mostly to heat or to some form of potential energy?

Step-by-Step Solution

Answer

The kinetic energy of the baseball when thrown by a major league pitcher at 150 km/h is approximately 125.4 Joules. When the speed of the baseball is decreased to 90 km/h, the kinetic energy decreases by a factor of approximately 0.361. When the baseball is caught by the catcher, the kinetic energy is mostly converted into heat, with a small portion being stored as potential energy.

Step 1: Write the formula for kinetic energy

Recall the formula for kinetic energy (KE): \(KE = \frac{1}{2}mv^2\) where m is the mass of the object and v is its velocity.

Step 2: Convert the given mass and speed of the baseball to appropriate units

Given mass m of the baseball is 145.4 g. We need to convert it to kilograms (kg): \(\displaystyle\frac{145.4 \mathrm{~g}}{1000\mathrm{~g/kg}} = 0.1454 \mathrm{~kg}\) Given speed v of the baseball is 150 km/h. We need to convert it to meters per second (m/s): \(\displaystyle\frac{150\;\mathrm{km/h}}{3.6\;\mathrm{m/s\cdot km/h}} = 41.67 \;\mathrm{m/s}\)

Step 3: Calculate the kinetic energy of the baseball

Now substitute the values of mass and velocity we just obtained into the kinetic energy formula: \(KE = \frac{1}{2}(0.1454\;\mathrm{kg})(41.67\;\mathrm{m/s})^2\) Upon calculating, \(KE \approx 125.402\;\mathrm{J}\) Thus, the kinetic energy of the baseball when thrown is approximately 125.4 Joules.

Step 4: Determine the factor by which the kinetic energy changes

When the speed of the baseball is decreased to 90 km/h, first convert it to m/s: \(\displaystyle\frac{90\;\mathrm{km/h}}{3.6\;\mathrm{m/s\cdot km/h}} = 25\;\mathrm{m/s}\) Next, calculate the new kinetic energy (KE') at this velocity: \(KE' = \frac{1}{2}(0.1454\;\mathrm{kg})(25\;\mathrm{m/s})^2 \approx 45.235\;\mathrm{J}\) Now we find the factor by which the kinetic energy changed: \(\displaystyle Factor = \frac{KE'}{KE} = \frac{45.235\;\mathrm{J}}{125.402\;\mathrm{J}}\) Calculating the factor, we get: \(Factor \approx 0.3609\) Therefore, the kinetic energy decreases by a factor of approximately 0.361 when the speed of the baseball is decreased to 90 km/h.

Step 5: Discuss the conversion of kinetic energy when caught by the catcher

When the baseball is caught by the catcher, the kinetic energy is mostly converted into heat and a small portion may be stored as potential energy (if the catcher's hand moves backward slightly). This is because the deceleration of the baseball when caught generates heat due to friction between the ball and the mitt, and much of the kinetic energy is absorbed by the catcher's hand and glove in the form of heat energy. The potential energy is relatively small and negligible compared to the heat generated.

Key Concepts

Mass Conversion

When calculating kinetic energy, having consistent units is crucial. In physics, we often use kilograms for mass. To convert grams to kilograms, we divide the mass value by 1000. For example, if a baseball weighs 145.4 grams, it should be converted to kilograms by calculating \[ \frac{145.4\,\text{g}}{1000\,\text{g/kg}} = 0.1454\,\text{kg} \].
This conversion ensures that we are using the correct units in all physics equations. Such conversions prevent errors and make calculations more straightforward. The use of kilograms is standard because it simplifies working with other metric units in formulas like those for kinetic energy.

Velocity Conversion

Velocity is another pivotal element when calculating kinetic energy. Often, speeds are given in kilometers per hour, especially in real-world scenarios. For kinetic energy calculations, converting this to meters per second is essential. The conversion factor is that 1 km/h equals 0.27778 m/s, which simplifies to dividing by 3.6.
For example, a speed of 150 km/h should be converted to:\[ \frac{150\,\text{km/h}}{3.6} = 41.67\,\text{m/s} \].This conversion aligns with the International System of Units (SI units) that the kinetic energy formula uses. Converting to m/s ensures the velocity is compatible with mass in kilograms and other SI units used in physics.

Energy Transformation

Energy transformation involves converting energy from one form to another. In kinetic energy problems, transformations often occur when an object changes speed or is subject to external forces. For example, when a baseball is caught, its kinetic energy (energy of motion) is transformed.
Most of this transformation results in heat energy due to friction as the glove decelerates the ball. A small amount might convert to potential energy if there's a displacement when the catcher's glove moves. Understanding these transformations is key in physics, as it illustrates energy conservation and transformation principles. Noting which energy form prevails helps in visualizing energy flow in physical processes.

Physics Problems

Physics problems are often tackled step-by-step, using known equations and methods. For calculating kinetic energy, follow these guidelines:

• Identify the given mass and velocity in the problem.
• Convert these values to the appropriate units (kilograms and meters per second) using conversion factors.
• Substitute these converted values into the kinetic energy formula, \( KE = \frac{1}{2}mv^2 \), to calculate the energy.
• Analyze the outcomes, such as changes in energy with altered velocities.
• Consider what happens to energy in the context of the problem, like what energy forms result from interactions and whether they align with energy conservation principles.

By addressing these steps meticulously, complex problems become manageable and clear. This systematic approach is invaluable in both academic exercises and real-world applications.