Problem 31
Question
A small car is traveling at twice the speed of a larger car, which has twice the mass of the smaller car. Which car has the greater kinetic energy? (Or do they both have the same kinetic energy?)
Step-by-Step Solution
Verified Answer
The small car has the greater kinetic energy.
1Step 1: Understanding Kinetic Energy Formula
Kinetic energy (KE) of an object is given by the formula: \[ KE = \frac{1}{2}mv^2 \]where \( m \) is the mass and \( v \) is the velocity of the object. Our goal is to compare the kinetic energies of two cars.
2Step 2: Defining Variables
Let the mass of the smaller car be \( m \) and its speed be \( v \). Thus, the mass of the larger car will be \( 2m \) and its speed will be \( \frac{v}{2} \).
3Step 3: Calculating Kinetic Energy of Smaller Car
Using the formula \( KE = \frac{1}{2}mv^2 \), we substitute \( m \) for the mass and \( v \) for the speed:\[ KE_{small} = \frac{1}{2}m(v)^2 = \frac{1}{2}mv^2 \]
4Step 4: Calculating Kinetic Energy of Larger Car
For the larger car, substitute \( 2m \) for the mass and \( \frac{v}{2} \) for the speed:\[ KE_{large} = \frac{1}{2}(2m)\left(\frac{v}{2}\right)^2 \]
5Step 5: Simplifying Kinetic Energy of Larger Car
Continuing from the expression for the larger car:\[ KE_{large} = \frac{1}{2} \times 2m \times \frac{v^2}{4} = \frac{1}{2} \times \frac{2m}{4} \times v^2 = \frac{1}{4}mv^2 \]
6Step 6: Comparison of Kinetic Energies
Now, compare \( KE_{small} = \frac{1}{2}mv^2 \) with \( KE_{large} = \frac{1}{4}mv^2 \). Clearly, \( \frac{1}{2}mv^2 > \frac{1}{4}mv^2 \). Thus, the smaller car has the greater kinetic energy.
Key Concepts
Kinetic Energy FormulaPhysics ProblemsMass and Velocity in Kinetic EnergyComparing Kinetic Energies
Kinetic Energy Formula
In physics, understanding the Kinetic Energy Formula is crucial. Kinetic energy measures the energy an object possesses due to its motion. The formula is represented as \( KE = \frac{1}{2}mv^2 \), where \( KE \) stands for kinetic energy, \( m \) is the mass of the object, and \( v \) is its velocity. This formula shows that kinetic energy is directly proportional to the mass of the object and the square of its velocity.
- If the velocity doubles, the kinetic energy quadruples, because velocity is squared.
- If the mass doubles, the kinetic energy doubles.
Physics Problems
Physics problems often require the application of formulas to find unknown quantities or to compare different scenarios. In the case of kinetic energy, problems typically involve identifying the mass and velocity of objects. Then, using the Kinetic Energy Formula to compute their energies.
Examples might include:
Examples might include:
- Comparing the kinetic energies of two vehicles due to changes in mass and velocity.
- Solving for one of the variables (mass or velocity) when the kinetic energy is given.
Mass and Velocity in Kinetic Energy
When dealing with kinetic energy, two critical factors are mass and velocity. Mass refers to how much matter is in an object, while velocity refers to the speed and direction that object is traveling. In terms of kinetic energy, even small changes in velocity have a significant impact because velocity is squared in the formula. This means:
- A small increase in velocity results in a much larger increase in kinetic energy compared to the same percentage change in mass.
- For two objects with the same velocity, the one with larger mass will have more kinetic energy.
Comparing Kinetic Energies
Comparing kinetic energies between two objects requires careful consideration of their masses and velocities. For instance, in the problem involving two cars, one smaller car travels faster (twice the speed), while the other is heavier (double the mass). By applying the Kinetic Energy Formula, we can determine:
- The smaller car, despite its lesser mass, has a greater kinetic energy because its speed has a much larger influence.
- Conversely, the larger car's slower speed significantly reduces its kinetic energy, more than its increased mass compensates for.
Other exercises in this chapter
Problem 29
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a. A \(100 .-\mathrm{g}\) sample of water is placed in an insulated container and allowed to come to room temperature at \(21^{\circ} \mathrm{C}\). To heat the
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A \(250-\mathrm{g}\) sample of water at \(20.0^{\circ} \mathrm{C}\) is placed in a freezer that is held at a constant temperature of \(-20.0^{\circ} \mathrm{C}\
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What is the enthalpy change for the preparation of one mole of liquid water from the elements, given the following equations? $$ \begin{aligned} &\mathrm{H}_{2}
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