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Chapter 5: Problem 13

(a) Calculate the kinetic energy in joules of a \(45-\mathrm{g}\) golf ball moving at \(61 \mathrm{~m} / \mathrm{s}\). (b) Convert this energy to calories. (c) What happens to this energy when the ball lands in a sand trap?

Short Answer

Expert verified
The kinetic energy of a \(45g\) golf ball moving at \(61 m/s\) is \(83.8 J\). When converted to calories, this is \(20 cal\). When the golf ball lands in a sand trap, its kinetic energy is transformed into other forms of energy, such as heat and potential energy, as it displaces the sand and comes to rest.

Step by step solution

01

Calculate the kinetic energy in joules

To calculate the kinetic energy, we can use the following formula: \[K.E. = \frac{1}{2}mv^2\] where \(K.E.\) is the kinetic energy, \(m\) is the mass of the golf ball, and \(v\) is its velocity. Given that the mass of the golf ball is \(45 g\) and its velocity is \(61 m/s\), we need to convert the mass unit from grams to kilograms: \[45 g = 0.045 kg\] Now, plug the mass and velocity into the formula: \[K.E. = \frac{1}{2}(0.045 kg)(61 m/s)^2\]
02

Calculate the kinetic energy

Calculate the kinetic energy using the values given: \[K.E. = \frac{1}{2}(0.045 kg)(61 m/s)^2 = 83.8 J\] So, the kinetic energy of the golf ball is \(83.8 J\).
03

Convert the kinetic energy to calories

To convert the kinetic energy from joules to calories, we can use the following conversion factor: \[1 cal = 4.184 J\] Therefore, the energy in calories is: \[\frac{83.8 J}{4.184 J/cal} = 20 cal\] So, the kinetic energy of the golf ball is \(20 cal\).
04

Explain what happens to the energy when the ball lands in a sand trap

When the golf ball lands in a sand trap, the kinetic energy is transformed into other forms of energy. Some of the energy will be used to displace the sand as the ball impacts and sinks into it. The energy is converted to heat due to friction between the golf ball and the sand particles, as well as potential energy as the sand particles move and rearrange themselves. Ultimately, the energy is dissipated, and the ball comes to rest in the sand trap.

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Key Concepts

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

Unit Conversion
Unit conversion is an essential tool in solving physics problems, especially when calculating kinetic energy. In the given exercise, the mass of the golf ball is given in grams, but the kinetic energy formula requires us to use kilograms. To convert grams to kilograms, the conversion factor is:
  • 1 kilogram = 1000 grams
Thus, to convert 45 grams to kilograms, you divide by 1000:
45 g = 45 / 1000 = 0.045 kg
This conversion ensures the correct units are used in the kinetic energy calculation. Unit conversions allow for consistency in calculations and help avoid errors when using formulas that assume specific units.
Energy Transformation
Energy transformation occurs when energy changes from one form to another. In the context of the golf ball, when it lands in a sand trap, its kinetic energy is not lost but transformed. Initially, the ball's kinetic energy is due to its motion.
Upon impact with the sand, this energy is primarily converted into:
  • Heat: Generated through friction between the ball and sand particles.
  • Sound: Produced from the impact noise.
  • Potential energy: As sand particles are displaced and rearranged.
This transformation illustrates the principle of conservation of energy, where energy is neither created nor destroyed, but it changes forms when an object interacts with its environment.
Physics Concepts
Several core physics concepts are at play in the kinetic energy calculation exercise. Understanding these can aid in comprehending the problem's mechanics:
  • Conservation of Energy: Energy remains constant in an isolated system, merely transforming between types.
  • Mass and Velocity: Both are crucial in calculating kinetic energy, emphasizing their roles in an object's motion.
  • Friction: This force opposes motion, converting kinetic energy into heat during the collision in the sand trap.
These concepts highlight the interconnectedness of various physical phenomena and reinforce basic principles studied in physics.
Kinetic Energy Formula
The kinetic energy ( K.E.) of an object can be determined using the formula:
\[K.E. = \frac{1}{2}mv^2\]Here, \(m\) represents mass, and \(v\) stands for velocity. The formula illustrates that kinetic energy is directly proportional to both mass and the square of velocity. This means:
  • Doubling the mass will double the kinetic energy.
  • Doubling the velocity quadruples the kinetic energy.
When calculating kinetic energy, it's vital to ensure correct unit usage, keeping mass in kilograms and velocity in meters per second to produce energy in joules. This formula is a cornerstone of basic physics, articulating the energy associated with motion.

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Most popular questions from this chapter

(a) Why is the change in enthalpy usually easier to measure than the change in internal energy? (b) For a given process at constant pressure, \(\Delta H\) is negative. Is the process endothermic or exothermic?

At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat \(\mathrm{KClO}_{3}\) : \(2 \mathrm{KClO}_{3}(s) \rightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g) \quad \Delta H=-89.4 \mathrm{~kJ}\) For this reaction, calculate \(\Delta H\) for the formation of (a) \(0.632 \mathrm{~mol}\) of \(\mathrm{O}_{2}\) and (b) \(8.57 \mathrm{~g}\) of \(\mathrm{KCl}\). (c) The decomposition of \(\mathrm{KClO}_{3}\) proceeds spontaneously when it is heated. Do you think that the reverse reaction, the formation of \(\mathrm{KClO}_{3}\) from \(\mathrm{KCl}\) and \(\mathrm{O}_{2}\), is likely to be feasible under ordinary conditions? Explain your answer.

The sun supplies about \(1.0\) kilowatt of energy for each square meter of surface area \(\left(1.0 \mathrm{~kW} / \mathrm{m}^{2}\right.\), where a watt \(=1 \mathrm{~J} / \mathrm{s})\). Plants produce the equivalent of about \(0.20 \mathrm{~g}\) of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) per hour per square meter. Assuming that the sucrose is produced as follows, calculate the percentage of sunlight used to produce sucrose. $$ \begin{aligned} 12 \mathrm{CO}_{2}(g)+11 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+12 \mathrm{O}_{2}(g) \\ \Delta H &=5645 \mathrm{~kJ} \end{aligned} $$

The air bags that provide protection in autos in the event of an accident expand because of a rapid chemical reaction. From the viewpoint of the chemical reactants as the system, what do you expect for the signs of \(q\) and \(w\) in this process?

Consider the following reaction, which occurs at room temperature and pressure: $$ 2 \mathrm{Cl}(g) \longrightarrow \mathrm{Cl}_{2}(g) \quad \Delta H=-243.4 \mathrm{~kJ} $$ Which has the higher enthalpy under these conditions, \(2 \mathrm{Cl}(g)\) or \(\mathrm{Cl}_{2}(g) ?\)
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