Problem 123
Question
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 \(7.87 \mathrm{~g} / \mathrm{cm}^{3} .\)
Step-by-Step Solution
Verified Answer
The velocity is approximately 23.58 meters per second.
1Step 1: 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 \( v \).
2Step 2: Convert Mass to Kilograms
The mass is given as \(56.6\mathrm{~g}\). To convert it to kilograms, divide by 1000. Therefore, \( m = \frac{56.6}{1000} = 0.0566 \mathrm{~kg}\).
3Step 3: Re-arrange the Kinetic Energy Formula
To solve for \( v \), re-arrange the formula \( KE = \frac{1}{2}mv^2 \) to \( v^2 = \frac{2 \cdot KE}{m} \).
4Step 4: 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 \( v^2 = \frac{31.5}{0.0566} \).
5Step 5: Calculate \( v^2 \) and \( v \)
Calculate \( v^2 = 556.01 \), and then take the square root to find \( v \). Thus, \( v \approx 23.58 \mathrm{~m/s} \).
Key Concepts
Mass ConversionVelocity CalculationProblem Solving Steps
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.
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.
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: \( v^2 = \frac{2 \cdot KE}{m} \)
- \( KE = 15.75 \) Joules
- \( m = 0.0566 \) kg
- Therefore, \( v^2 = \frac{31.5}{0.0566} \)
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.
Other exercises in this chapter
Problem 121
Describe the physical characteristics of white phosphorus. Is it found in any modern matches? Why or why not?
View solution Problem 122
What is the phosphorus compound used in "strike anywhere" matches. What is the chemical equation for the burning of this compound in air?
View solution Problem 124
Sulfur dioxide gas reacts with oxygen, \(\mathrm{O}_{2}(\mathrm{~g})\), to produce \(\mathrm{SO}_{3}(g)\). This reaction releases \(99.0 \mathrm{~kJ}\) of heat
View solution Problem 126
How many grams of oxygen gas are required to produce \(7.60 \mathrm{~kJ}\) of heat when hydrogen gas burns at constant pressure to produce gaseous water? $$ 2 \
View solution