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Which has more kinetic energy, a \(1400-\mathrm{kg}\) car moving at \(115 \mathrm{~km} / \mathrm{h}\) or a \(12,000-\mathrm{kg}\) truck moving at \(38 \mathrm{~km} / \mathrm{h} ?\)

Short Answer

Expert verified
The car has more kinetic energy.

Step by step solution

01

Understand the Kinetic Energy Formula

Kinetic energy (KE) is given by the formula: \(KE = \frac{1}{2}mv^2\), where \(m\) is the mass in kilograms, and \(v\) is the velocity in meters per second (m/s). We first need to ensure all velocities are in m/s.
02

Convert Velocity Units

Convert the velocities from km/h to m/s by using the conversion factor: \(1 \text{ km/h} = \frac{1}{3.6} \text{ m/s}\). For the car: \(v = 115 \times \frac{1}{3.6} \approx 31.94\) m/s. For the truck: \(v = 38 \times \frac{1}{3.6} \approx 10.56\) m/s.
03

Calculate Kinetic Energy of the Car

Substitute the car's mass and velocity into the kinetic energy formula: \(KE_{\text{car}} = \frac{1}{2} \times 1400 \times (31.94)^2\). Calculate this to find \(KE_{\text{car}}\).
04

Calculate Kinetic Energy of the Truck

Substitute the truck's mass and velocity into the kinetic energy formula: \(KE_{\text{truck}} = \frac{1}{2} \times 12000 \times (10.56)^2\). Calculate this to find \(KE_{\text{truck}}\).
05

Compare the Kinetic Energies

Compare the values of \(KE_{\text{car}}\) and \(KE_{\text{truck}}\) to determine which is greater.

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Key Concepts

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

Understanding the Kinetic Energy Formula
The kinetic energy formula is a fundamental concept in physics, allowing us to determine the energy possessed by a moving object. The formula is expressed as: \[ KE = \frac{1}{2} mv^2 \] where:
  • \( KE \) represents kinetic energy measured in Joules ( \( J \)).
  • \( m \) refers to mass in kilograms ( \( kg \)).
  • \( v \) is velocity in meters per second ( \( m/s \)).
The key takeaway is that kinetic energy depends on both mass and the square of velocity. This means that changes in velocity have a more significant impact on kinetic energy compared to changes in mass.
Role of Mass in Kinetic Energy
Mass is a measure of the amount of matter in an object and it plays a crucial role in calculating kinetic energy. Even though both mass and velocity impact kinetic energy, it's important to understand how mass influences this value.
  • Doubling the mass while keeping the velocity constant will double the kinetic energy.
  • Mass is directly proportional to kinetic energy, meaning they increase or decrease together.
For instance, in the previous exercise, we compared a car and a truck. Despite the truck having a larger mass, its kinetic energy also depends heavily on its velocity.
Converting Velocity Units for Consistency
To accurately calculate kinetic energy, velocity must be in the correct unit: meters per second (m/s). In practical exercises, velocities are often given in kilometers per hour (km/h). To convert these units, use the conversion factor:\[ 1 \text{ km/h} = \frac{1}{3.6} \text{ m/s} \]This conversion ensures that you can apply the kinetic energy formula correctly. In the example given:
  • The car's velocity of 115 km/h converts to roughly 31.94 m/s.
  • The truck's velocity of 38 km/h converts to approximately 10.56 m/s.
Always double-check your conversions to avoid errors in the kinetic energy calculations.
Comparing Energy of Different Objects
Once kinetic energies are calculated for different objects, you can compare them to see which has more energy. Understanding this comparison helps us make sense of how mass and velocity together influence energy.
  • A larger velocity generally results in more kinetic energy, given the quadratic nature of the velocity term in the formula.
  • A larger mass can also lead to higher energy, but the impact is linear compared to the velocity.
In comparing the car and truck from the exercise, the calculations show differing results primarily due to the velocity of each vehicle. While the truck has more mass, the car's higher velocity plays a crucial role in the kinetic energy outcome. These comparisons are useful in many fields like automotive safety, engineering, and even sports.

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Most popular questions from this chapter

When a sample of a hydrocarbon fuel is ignited and burned in oxygen, the internal energy decreases by \(7.20 \mathrm{~kJ} .\) If \(5670 \mathrm{~J}\) of heat were transferred to the surroundings, what is the sign and magnitude of work? If the reaction took place in an environment with a pressure of \(1 \mathrm{~atm}\), what was the volume change?

Acetic acid \(\left(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right)\), whose aqueous solutions are known as vinegar, is prepared by reaction of ethyl alcohol ( \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) ) with oxygen: $$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(l)+\mathrm{H}_{2} \mathrm{O}(l)$$ Use the following data to calculate \(\Delta H^{\circ}\) in kilojoules for the reaction: $$\begin{aligned}&\Delta H_{\mathrm{f}}\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(l)\right]=-277.7 \mathrm{~kJ} / \mathrm{mol} \\ &\Delta H^{\circ} \mathrm{f}\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(l)\right]=-484.5 \mathrm{~kJ} / \mathrm{mol} \\ &\Delta H_{\mathrm{f}}^{\circ}\left[\mathrm{H}_{2} \mathrm{O}(l)\right]=-285.8 \mathrm{~kJ} / \mathrm{mol} \end{aligned}$$

Chloroform has \(\Delta H_{\text {vaporimation }}=29.2 \mathrm{~kJ} / \mathrm{mol}\) and boils at \(61.2{ }^{\circ} \mathrm{C}\). What is the value of \(\Delta S_{\text {vaporization }}\) for chloroform?

Assume that the kinetic energy of a \(1400-\mathrm{kg}\) car moving at \(115 \mathrm{~km} / \mathrm{h}\) (Problem \(9.42\) ) could be converted entirely into heat. What amount of water could be heated from \(20^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\) by the car's energy? \(4.184 \mathrm{~J}\) are required to heat \(1 \mathrm{~g}\) of water by \(1{ }^{\circ} \mathrm{C}\).

What is a compound's standard heat of formation?

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