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Chapter 1: Problem 26

(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?

Short Answer

Expert verified
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 by step solution

01

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.
02

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): \(\displaystyle\frac{150\;\mathrm{km/h}}{3.6\;\mathrm{m/s\cdot km/h}} = 41.67 \;\mathrm{m/s}\)
03

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.
04

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.
05

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.

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