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Chapter 6: Problem 123

How fast (in meters per second) must an iron ball with a mass of \(56.6 \mathrm{~g}\) be traveling in order to have a kinetic energy of \(15.75 \mathrm{~J} ?\) The density of iron is \(7.87 \mathrm{~g} / \mathrm{cm}^{3} .\)

Short Answer

Expert verified
The velocity is approximately 23.58 meters per second.

Step by step solution

01

Understand the Kinetic Energy Formula

The formula for kinetic energy is given by \( KE = \frac{1}{2}mv^2 \), where \( KE \) is the kinetic energy, \( m \) is mass in kilograms, and \( v \) is velocity in meters per second. Our goal is to solve for velocity \( v \).
02

Convert Mass to Kilograms

The mass is given as \(56.6\mathrm{~g}\). To convert it to kilograms, divide by 1000. Therefore, \( m = \frac{56.6}{1000} = 0.0566 \mathrm{~kg}\).
03

Re-arrange the Kinetic Energy Formula

To solve for \( v \), re-arrange the formula \( KE = \frac{1}{2}mv^2 \) to \( v^2 = \frac{2 \cdot KE}{m} \).
04

Plug in the Known Values

Substitute \( KE = 15.75 \mathrm{~J} \) and \( m = 0.0566 \mathrm{~kg} \) into the equation \( v^2 = \frac{2 \cdot 15.75}{0.0566} \), giving \( v^2 = \frac{31.5}{0.0566} \).
05

Calculate \( v^2 \) and \( v \)

Calculate \( v^2 = 556.01 \), and then take the square root to find \( v \). Thus, \( v \approx 23.58 \mathrm{~m/s} \).

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Key Concepts

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

Mass Conversion
When dealing with physics problems, it's crucial to use consistent units. In the original exercise, the mass of the iron ball is provided as 56.6 grams. However, in physics, the standard unit of mass is kilograms.
To convert grams to kilograms, which is the required unit for the kinetic energy formula, divide the mass in grams by 1000.
  • The given mass is 56.6 grams.
  • To convert to kilograms, calculate: \( m = \frac{56.6}{1000} = 0.0566 \) kg.
This conversion is essential as the kinetic energy formula requires mass in kilograms. Always remember to check your units before solving the problem!
Velocity Calculation
The task is to find the velocity of the iron ball given its mass and kinetic energy. We begin with the kinetic energy formula: \( KE = \frac{1}{2} mv^2 \). Here, our objective is to isolate and solve for velocity \( v \).First, rearrange the equation to make \( v \) the subject:
  • We know: \( KE = \frac{1}{2} mv^2 \)
  • Re-arranging gives: \( v^2 = \frac{2 \cdot KE}{m} \)
Once rearranged, substitute the known values into this equation. For the given problem:
  • \( KE = 15.75 \) Joules
  • \( m = 0.0566 \) kg
  • Therefore, \( v^2 = \frac{31.5}{0.0566} \)
Calculate \( v^2 \) and take the square root to find the velocity \( v \). The final step gives us approximately 23.58 \( \mathrm{m/s} \). This way, you can determine the speed required for a given kinetic energy.
Problem Solving Steps
When tackling physics problems, breaking things into clear, manageable steps is key. This helps avoid mistakes and ensures each part of the problem is properly addressed.Here's how the problem was approached in the solution:
  • Step 1: Understand the given formula - In this case, \( KE = \frac{1}{2} mv^2 \). Recognize what you need to find and what you have.
  • Step 2: Convert necessary values to suitable units, such as converting grams to kilograms.
  • Step 3: Re-arrange the formula to isolate the variable you are solving for, namely velocity \( v \).
  • Step 4: Substitute all known values into the equation.
  • Step 5: Perform calculations to solve for the unknown and check your work for accuracy.
By following such a structured approach, solving physics problems becomes easier and more comprehensible. Each step logically leads to the next, creating a clear path to the solution.

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Most popular questions from this chapter

When steam condenses to liquid water, \(2.26 \mathrm{~kJ}\) of heat is released per gram. The heat from \(168 \mathrm{~g}\) of steam is used to heat a room containing \(6.44 \times 10^{4} \mathrm{~g}\) of air \((20 \mathrm{ft} \times 12 \mathrm{ft} \times 8 \mathrm{ft})\). The specific heat of air at normal pressure is \(1.015 \mathrm{~J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right) .\) What is the change in air temperature, assuming the heat from the steam is all absorbed by air?

Carbon disulfide burns in air, producing carbon dioxide and sulfur dioxide. $$ \mathrm{CS}_{2}(l)+3 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{SO}_{2}(g) ; \Delta H=-1077 \mathrm{~kJ} $$ What is \(\Delta H\) for the following equation? $$ \frac{1}{2} \mathrm{CS}_{2}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \frac{1}{2} \mathrm{CO}_{2}(g)+\mathrm{SO}_{2}(g) $$

Part 1: In an insulated container, you mix 200. g of water at \(80^{\circ} \mathrm{C}\) with \(100 . \mathrm{g}\) of water at \(20^{\circ} \mathrm{C}\). After mixing, the temperature of the water is \(60^{\circ} \mathrm{C}\). a. How much did the temperature of the hot water change? How much did the temperature of the cold water change? Compare the magnitudes (positive values) of these changes. b. During the mixing, how did the heat transfer occur: from hot water to cold, or from cold water to hot? C. What quantity of heat was transferred from one sample to the other? d. How does the quantity of heat transferred to or from the hot-water sample compare with the quantity of heat transferred to or from the cold-water sample? e. Knowing these relative quantities of heat, why is the temperature change of the cold water greater than the magnitude of the temperature change of the hot water. f. A sample of hot water is mixed with a sample of cold water that has twice its mass. Predict the temperature change of each of the samples. g. You mix two samples of water, and one increases by \(20^{\circ} \mathrm{C}\), while the other drops by \(60^{\circ} \mathrm{C}\). Which of the samples has less mass? How do the masses of the two water samples compare? h. A 7 -g sample of hot water is mixed with a \(3-\mathrm{g}\) sample of cold water. How do the temperature changes of the two water samples compare? Part \(2:\) A sample of water is heated from \(10^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\). Can you calculate the amount of heat added to the water sample that caused this temperature change? If not, what information do you need to perform this calculation? Part 3: Two samples of water are heated from \(20^{\circ} \mathrm{C}\) to \(60^{\circ} \mathrm{C}\). One of the samples requires twice as much heat to bring about this temperature change as the other. How do the masses of the two water samples compare? Explain your reasoning.

Calcium oxide (quicklime) reacts with water to produce calcium hydroxide (slaked lime). $$ \mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s) ; \Delta H=-65.2 \mathrm{~kJ} $$ The heat released by this reaction is sufficient to ignite paper. How much heat is released when \(24.5 \mathrm{~g}\) of calcium oxide reacts?

Under what condition is the enthalpy change equal to the heat of reaction?
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