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Chapter 1: Problem 26

(a) A baseball weighs 5.13 oz. What is the kinetic energy, in joules, of this baseball when it is thrown by a major league pitcher at 95.0 \(\mathrm{mi} / \mathrm{h} ?\) (b) By what factor will the kinetic energy change if the speed of the baseball is decreased to 55.0 \(\mathrm{mi} / \mathrm{h} ?\) (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
(a) The kinetic energy of the baseball thrown at 95.0 mi/h is 132.318 J. (b) The kinetic energy changes by a factor of approximately 0.331 when the speed is decreased to 55.0 mi/h. (c) When the baseball is caught, most of the kinetic energy is converted into heat due to deformation and friction; a small fraction might be temporarily converted into potential energy but will ultimately dissipate as heat.

Step by step solution

01

(Step 1: Convert units to SI )

As we need to find the kinetic energy in joules, we need to make sure that the mass and velocity are in SI units, i.e., kg and m/s, respectively. First, we convert the mass of the baseball from ounces (oz) to kilograms (kg) and the velocity from mi/h to m/s. Conversion factors: 1 oz = 0.0283495 kg 1 mi/h = 0.44704 m/s \(mass = 5.13\,oz * \frac{0.0283495\,kg}{1\,oz} = 0.1454\,kg\) \(velocity_1 = 95.0\,mi/h * \frac{0.44704\,m/s}{1\,mi/h} = 42.47\,m/s\) \(velocity_2 = 55.0\,mi/h * \frac{0.44704\,m/s}{1\,mi/h} = 24.587\,m/s\)
02

(Step 2: Calculate the kinetic energy)

Now, we will use the formula for kinetic energy. Kinetic energy formula: \(KE = \frac{1}{2} mv^2\) (a) To find the initial kinetic energy of the baseball thrown at 95.0 mi/h, substitute the converted mass and velocity values into the kinetic energy formula. \(KE_1 = \frac{1}{2}(0.1454\,kg)(42.47\,m/s)^2\) \(KE_1 = 132.318\,J\) (b) To find the kinetic energy of the baseball when the speed decreases to 55.0 mi/h, repeat the process using the second velocity value. \(KE_2 = \frac{1}{2}(0.1454\,kg)(24.587\,m/s)^2\) \(KE_2 = 43.817\,J\) Now, to find the factor by which the kinetic energy changes, we divide the final kinetic energy by the initial kinetic energy. \(Factor = \frac{KE_2}{KE_1} = \frac{43.817\,J}{132.318\,J} = 0.331\) So, the kinetic energy changes by a factor of approximately 0.331. (c) When the baseball is caught, its velocity becomes zero. Since kinetic energy depends on the square of the velocity, the kinetic energy also becomes zero. This energy is mostly converted into heat due to deformation and friction between the ball and the catcher's glove. A small fraction of the energy may also be temporarily converted into potential energy if the catcher's hand moves backward upon catching the ball, but this energy will also dissipate as heat when the hand comes to rest.

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Key Concepts

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

SI Unit Conversion
Converting units to the International System of Units (SI) is a fundamental step in physics when working with equations and formulas. The SI unit for mass is the kilogram (kg), and for velocity, it is meters per second (m/s). In our exercise, the mass of a baseball was initially given in ounces and the velocity in miles per hour. To work with the kinetic energy formula, we need to convert these values to kilograms and meters per second.

Here's how the conversion factors apply:
  • 1 ounce is equivalent to 0.0283495 kilograms.
  • 1 mile per hour converts to 0.44704 meters per second.
Using these conversion factors, we can accurately transform measurements and then proceed to calculate the kinetic energy. Correct unit conversion is crucial because it ensures that the values computed are in the correct scale and dimension for the physical quantities we are exploring.
Kinetic Energy Formula
Kinetic energy is the energy that an object possesses due to its motion. The formula to calculate the kinetic energy (\(KE\)) of an object is: \[ KE = \frac{1}{2} mv^{2} \] where:
  • \(m\) represents the mass of the object in kilograms,
  • \(v\) represents the velocity of the object in meters per second,
and the factor of \(\frac{1}{2}\) is a constant that arises from the integration of the work-energy principle. Using this formula allows us to determine the amount of work that can be performed by an object in motion - or, equivalently, the kinetic energy it has due to its velocity. To find the kinetic energy of the baseball, we substituted the mass converted to kilograms and velocity converted to meters per second into the formula, yielding energy values in joules (\(J\)), the SI unit of energy.
Energy Conservation
The principle of energy conservation states that energy cannot be created or destroyed; it can only be transformed from one form to another. In the context of our baseball scenario, when a pitcher throws a ball, the muscular energy is converted into kinetic energy of the ball. As the ball is caught by the catcher, the kinetic energy is transformed into other forms of energy. Primarily, it turns into heat due to the friction and the deformation of the ball and the glove.

Some energy might be momentarily stored as potential energy if, for example, the catcher's mitt moves back upon impact. However, this potential energy is quickly converted back to heat when movement ceases. Understanding the transformations of energy forms helps us to comprehend how the kinetic energy of an object like a baseball changes as it interacts with its environment. It's crucial in physics to track energy flow because it reveals how different forms of energy contribute to a system's behavior over time.

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Most popular questions from this chapter

A solid white substance A is heated strongly in the absence of air. It decomposes to form a new white substance \(B\) and a gas C.The gas has exactly the same properties as the product obtained when carbon is burned in an excess of oxygen. Based on these observations, can we determine whether solids \(\mathrm{A}\) and \(\mathrm{B}\) and gas \(\mathrm{C}\) are elements or compounds?

A 40 -lb container of peat moss measures \(14 \times 20 \times 30\) in. A 40 -lb container of topsoil has a volume of 1.9 gal. (a) Calculate the average densities of peat moss and topsoil in units of\(\mathrm{g} / \mathrm{cm}^{3} .\) Would it be correct to say that peat moss is "lighter" than topsoil? (b) How many bags of peat moss are needed to cover an area measuring 15.0 \(\mathrm{ft} \times 20.0 \mathrm{ft}\) to a depth of 3.0 in.?

A thief plans to steal a gold sphere with a radius of 28.9 \(\mathrm{cm}\) from a museum. If the gold has a density of \(19.3 \mathrm{g} / \mathrm{cm}^{3},\) what is the mass of the sphere in pounds? [The volume of a sphere is \(V=(4 / 3) \pi r^{3} . ]\) Is the thief likely to be able to walk off with the gold sphere unassisted?

Use appropriate metric prefixes to write the following measurements without use of exponents: (a) \(2.3 \times 10^{-10} \mathrm{L}\) ,(b) \(4.7 \times 10^{-6} \mathrm{g},\) (c) \(1.85 \times 10^{-12} \mathrm{m},\) (d) \(16.7 \times 10^{6} \mathrm{s}\) (e) \(15.7 \times 10^{3} \mathrm{g},(\mathrm{f}) 1.34 \times 10^{-3} \mathrm{m},(\mathrm{g}) 1.84 \times 10^{2} \mathrm{cm}\)

A 25.0 -cm-long cylindrical glass tube, sealed at one end, is filled with ethanol. The mass of ethanol needed to fill the tube is found to be 45.23 g. The density of ethanol is 0.789 \(\mathrm{g} / \mathrm{mL}\) . Calculate the inner diameter of the tube in centimeters.
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