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Chapter 5: Problem 14

Suppose you toss a tennis ball upward. (a) Does the kinetic energy of the ball increase or decrease as it moves higher? (b) What happens to the potential energy of the ball as it moves higher? (c) If the same amount of energy were imparted to a ball the same size as a tennis ball but of twice the mass, how high would the ball go in comparison to the tennis ball? Explain your answers.

Short Answer

Expert verified
The kinetic energy of the tennis ball decreases as it moves higher due to the force of gravity slowing it down. The potential energy of the tennis ball increases as it moves higher since it depends on the height above the ground. When given the same amount of energy, the ball of twice the mass will reach half the height of the tennis ball as the initial energy gets divided by the increased mass.

Step by step solution

01

(Step 1: Determine the trend of Kinetic Energy as the ball moves higher)

As the tennis ball moves upward, it will gradually slow down due to the force of gravity acting on it. As the speed of the ball decreases, its kinetic energy will also decrease. Therefore, the kinetic energy of the tennis ball decreases as it moves higher.
02

(Step 2: Determine the trend of Potential Energy as the ball moves higher)

The potential energy of the tennis ball is determined by the formula \(PE = mgh\), where \(m\) is the mass of the ball, \(g\) is the acceleration due to gravity, and \(h\) is the height above the ground. As the height of the ball increases, the potential energy will also increase. Thus, the potential energy of the tennis ball increases as it moves higher.
03

(Step 3: Compare the heights reached by the tennis ball and a ball of twice its mass when given the same energy)

Let's consider a tennis ball of mass \(m\) and another ball of mass \(2m\). Suppose they receive the same amount of energy (\(E\)) when tossed upward. For the tennis ball, conservation of energy states that the initial energy should equal to the sum of the kinetic and potential energy when the ball reaches its highest point: \(E = mgh + \frac{1}{2}mv^2\) However, as the ball reaches its highest point, its velocity becomes zero as it momentarily stops for changing direction. Therefore, its kinetic energy becomes zero: \(E = mgh\) Now, let's calculate the potential energy of the ball with mass \(2m\), given the same energy \(E\): \(E = (2m)gh'\) As the energy is the same, we can compare the two expressions: \(mgh = (2m)gh'\), thus \(h = 2h'\), where \(h'\) is the height reached by the ball of mass \(2m\). Therefore, the ball of twice the mass will reach half the height of the tennis ball when given the same amount of energy.

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Key Concepts

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

Kinetic Energy
Let's explore kinetic energy, which is the energy an object possesses due to its motion. According to physics, the kinetic energy \( K \) of a moving object with mass \( m \) and velocity \( v \) is given by the formula \( K = \frac{1}{2}mv^2 \). What's fascinating is how kinetic energy changes as an object moves.

When you toss a tennis ball upward, the gravitational force slows it down until it reaches its peak height. During this climb, its speed鈥攁nd consequently, its kinetic energy鈥攄ecreases. This is why, just before the ball starts falling back down, it appears to hang motionless in the air; at that highest point, its kinetic energy is zero because its velocity is zero. To grasp this concept deeply, imagine a slow-motion video of the ball's ascent鈥攅ach frame shows the ball moving slower than the last, reflecting a continuous decrease in kinetic energy.
Potential Energy
Potential energy, on the other hand, is the stored energy of position possessed by an object. In the context of throwing a tennis ball into the air, we focus on gravitational potential energy, which depends on the object's height above the ground. The formula for potential energy \( PE \) is \( PE = mgh \), where \( m \) is mass, \( g \) is the acceleration due to gravity, and \( h \) is height.

As the ball rises, it trades kinetic energy for potential energy until, at the apex, all its energy is potential. This is a classic illustration of the conservation of energy, where the total energy remains constant but shifts between forms. Therefore, as you might have deduced, the potential energy of the tennis ball increases as it gains altitude, reaching a maximum when the ball reaches its highest point and comes momentarily to rest.
Gravitational Force
Gravitational force is a fundamental force that attracts two bodies with mass. On Earth, this force provides a constant acceleration downwards, at approximately \( 9.8 \text{ m/s}^2 \), designated as \( g \). It's this same force that acts on a ball thrown into the air, pulling it back down.

In the exercise provided, when comparing a tennis ball to a heavier ball tossed with the same energy, the gravitational force will affect the heavier ball more, as it has more mass. This means that a ball with twice the mass of a tennis ball will reach half the altitude, should the same energy be imparted to both. This outcome stems from the gravitational force acting on the kinetic and potential energies, influencing how high each ball will rise. Understanding how gravitational force interacts with mass and energy is crucial in analyzing movements not just on Earth, but in all sorts of celestial mechanics.

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Most popular questions from this chapter

(a) Why are tables of standard enthalpies of formation so useful? (b) What is the value of the standard enthalpy of formation of an element in its most stable form? (c) Write the chemical equation for the reaction whose enthalpy change is the standard enthalpy of formation of sucrose (table sugar), \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s), \Delta H_{f}^{\circ}\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right]\).

A \(1.800-g\) sample of phenol \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\right)\) was burned in a bomb calorimeter whose total heat capacity is \(11.66 \mathrm{~kJ} /{ }^{\circ} \mathrm{C}\). The temperature of the calorimeter plus contents increased from \(21.36\) to \(26.37^{\circ} \mathrm{C}\). (a) Write a balanced chemical equation for the bomb calorimeter reaction. (b) What is the heat of combustion per gram of phenol? Per mole of phenol?

Consider the following reaction: $$ 2 \mathrm{CH}_{3} \mathrm{OH}(g) \longrightarrow 2 \mathrm{CH}_{4}(g)+\mathrm{O}_{2}(g) \quad \Delta H=+252.8 \mathrm{~kJ} $$ (a) Is this reaction exothermic or endothermic? (b) Calculate the amount of heat transferred when \(24.0 \mathrm{~g}\) of \(\mathrm{CH}_{3} \mathrm{OH}(g)\) is decomposed by this reaction at constant pressure. (c) For a given sample of \(\mathrm{CH}_{3} \mathrm{OH}\), the enthalpy change during the reaction is \(82.1 \mathrm{~kJ}\). How many grams of methane gas are produced? (d) How many kilojoules of heat are released when \(38.5 \mathrm{~g}\) of \(\mathrm{CH}_{4}(g)\) reacts completely with \(\mathrm{O}_{2}(g)\) to form \(\mathrm{CH}_{3} \mathrm{OH}(\mathrm{g})\) at constant pressure? 5.45 When solutions containing silver ions and chloride ions are mixed, silver chloride precipitates: $$ \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \operatorname{AgCl}(s) \quad \Delta H=-65.5 \mathrm{~kJ} $$ (a) Calculate \(\Delta H\) for the production of \(0.450 \mathrm{~mol}\) of \(\mathrm{AgCl}\) by this reaction. (b) Calculate \(\Delta H\) for the production of \(9.00 \mathrm{~g}\) of \(\mathrm{AgCl}\). (c) Calculate \(\Delta H\) when \(9.25 \times 10^{-4} \mathrm{~mol}\) of \(\mathrm{AgCl}\) dissolves in water.

Complete combustion of \(1 \mathrm{~mol}\) of acetone \(\left(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}\right)\) liberates \(1790 \mathrm{~kJ}\) : $$ \begin{aligned} \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}(l)+4 \mathrm{O}_{2}(g) \longrightarrow 3 \mathrm{CO}_{2}(g)+& 3 \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H^{\circ}=-1790 \mathrm{~kJ} \end{aligned} $$ Using this information together with the standard enthalpies of formation of \(\mathrm{O}_{2}(g), \mathrm{CO}_{2}(g)\), and \(\mathrm{H}_{2} \mathrm{O}(l)\) from Appendix \(\mathrm{C}\), calculate the standard enthalpy of formation of acetone.

Consider the following reaction: $$ 2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s) \quad \Delta H=-1204 \mathrm{~kJ} $$ (a) Is this reaction exothermic or endothermic? (b) Calculate the amount of heat transferred when \(3.55 \mathrm{~g}\) of \(\mathrm{Mg}(s)\) reacts at constant pressure. (c) How many grams of \(\mathrm{MgO}\) are produced during an enthalpy change of \(-234 \mathrm{~kJ}\) ? (d) How many kilojoules of heat are absorbed when \(40.3 \mathrm{~g}\) of \(\mathrm{MgO}(s)\) is decomposed into \(\mathrm{Mg}(s)\) and \(\mathrm{O}_{2}(g)\) at constant pressure?
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