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Which has the greater kinetic energy, an object with a mass of \(2.0 \mathrm{kg}\) and a velocity of \(1.0 \mathrm{m} / \mathrm{s}\) or an object with a mass of \(1.0 \mathrm{kg}\) and a velocity of \(2.0 \mathrm{m} / \mathrm{s} ?\)

Short Answer

Expert verified
The object with a mass of 1.0 kg and a velocity of 2.0 m/s has a greater kinetic energy of 2.0 J, compared to the other object with 1.0 J.

Step by step solution

01

Identify the given values

We have two different objects given with specific masses and velocities: - Object 1: mass (m1) = 2.0 kg, velocity (v1) = 1.0 m/s - Object 2: mass (m2) = 1.0 kg, velocity (v2) = 2.0 m/s
02

Calculate the kinetic energy for Object 1

To calculate the kinetic energy for the first object, we will use the formula KE = (1/2)mv^2. Plugging in the values for m1 and v1, we get: KE1 = (1/2)*(2.0 kg)*(1.0 m/s)^2 KE1 = (1.0 kg)*(1.0 m^2/s^2) KE1 = 1.0 J (Joules)
03

Calculate the kinetic energy for Object 2

Similarly, we will calculate the kinetic energy for the second object using the mass (m2) and velocity (v2): KE2 = (1/2)*(1.0 kg)*(2.0 m/s)^2 KE2 = (0.5 kg)*(4.0 m^2/s^2) KE2 = 2.0 J (Joules)
04

Compare the kinetic energies of both objects

Now that we have the kinetic energies for both objects, we can easily compare them: - Object 1: KE1 = 1.0 J - Object 2: KE2 = 2.0 J Since 2.0 Joules is greater than 1.0 Joule, the object with a mass of 1.0 kg and a velocity of 2.0 m/s (Object 2) has the greater kinetic energy.

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Key Concepts

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

Kinetic Energy Formula
Kinetic energy is the energy that an object possesses due to its motion. To determine the kinetic energy of a moving object, a fundamental formula is used:
\[ KE = \frac{1}{2}mv^2 \]
where
  • \( KE \) is the kinetic energy in Joules (J),
  • \( m \) is the mass of the object in kilograms (kg), and
  • \( v \) is the velocity of the object in meters per second (m/s).

This equation implies that kinetic energy is directly proportional to the mass of the object and the square of its velocity. The factor of \( \frac{1}{2} \) in the formula is a constant that arises from the work-energy principle, showing that kinetic energy is a scalar quantity: it has magnitude but no direction. Understanding this formula is crucial for solving problems related to kinetic energy, as it allows you to calculate the energy for any object in motion given its mass and velocity.
Mass and Velocity in Kinetic Energy
The relationship between mass and velocity in kinetic energy is pivotal. When we look at the formula \[ KE = \frac{1}{2}mv^2 \], it reveals important insights:
  • Doubling the mass \( (m) \) of an object will double its kinetic energy, keeping velocity constant, because mass is directly proportional to kinetic energy.
  • Doubling the velocity \( (v) \) of an object will quadruple its kinetic energy since velocity is squared in the equation. Thus, velocity has a more dramatic effect on kinetic energy than mass does.

Consider an example where two balls of different masses are thrown at the same speed; the heavier ball will have more kinetic energy because of its greater mass. Conversely, if the two balls have the same mass but are thrown at different speeds, the one with the higher speed will possess significantly more kinetic energy. It's a delicate balance that is vital when comparing energies of moving objects, as the influence of mass and velocity on the kinetic energy needs to be carefully evaluated.
Comparing Kinetic Energy
When comparing the kinetic energy of different objects, such as in the presented exercise, two factors must be considered: mass and velocity. The challenge in comparing kinetic energy between objects comes from the fact that an increase in either mass or velocity will result in an increase in kinetic energy, but velocity has a greater impact because it is squared in the formula.
In the provided problem, Object 1 has a mass of \(2.0 \text{ kg}\) and a velocity of \(1.0 \text{ m/s}\), while Object 2 has a mass of \(1.0 \text{ kg}\) and a velocity of \(2.0 \text{ m/s}\). After calculating their kinetic energies, we establish that Object 2, despite having half the mass, has double the kinetic energy of Object 1. This result underscores the point that, when velocities are different, velocity can greatly outstrip mass in contributing to kinetic energy.
Simultaneously, it's useful to remember when comparing kinetic energies that identical masses with differing velocities or identical velocities with differing masses result in differing kinetic energies. Mastery of these concepts can greatly facilitate understanding and solving physics problems involving kinetic energy.

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Most popular questions from this chapter

Given the following data $$\begin{aligned}\mathrm{Ca}(s)+2 \mathrm{C}(\text {graphite}) & \longrightarrow \mathrm{CaC}_{2}(s) & & \Delta H=-62.8 \mathrm{kJ} \\\ \mathrm{Ca}(s)+\frac{1}{2} \mathrm{O}_{2}(g) & \longrightarrow \mathrm{CaO}(s) & & \Delta H=-635.5 \mathrm{kJ} \\ \mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q) &\Delta H &=-653.1 \mathrm{kJ} \\\ \mathrm{C}_{2} \mathrm{H}_{2}(g)+\frac{5}{2} \mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \Delta H &=-1300 . \mathrm{kJ} \\ \mathrm{C}(\text {graphite})+\mathrm{O}_{2}(g) & \longrightarrow \mathrm{CO}_{2}(\mathrm{g}) & \Delta H &=-393.5 \mathrm{kJ} \end{aligned}$$ calculate \(\Delta H\) for the reaction $$\mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{C}_{2} \mathrm{H}_{2}(g)$$

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Consider the following changes: a. \(\mathrm{N}_{2}(g) \longrightarrow \mathrm{N}_{2}(l)\) b. \(\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g)\) c. \(\mathrm{Ca}_{3} \mathrm{P}_{2}(s)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 3 \mathrm{Ca}(\mathrm{OH})_{2}(s)+2 \mathrm{PH}_{3}(g)\) d. \(2 \mathrm{CH}_{3} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)\) e. \(\mathrm{I}_{2}(s) \longrightarrow \mathrm{I}_{2}(g)\) At constant temperature and pressure, in which of these changes is work done by the system on the surroundings? By the surroundings on the system? In which of them is no work done?

A \(150.0 -\mathrm{g}\) sample of a metal at \(75.0^{\circ} \mathrm{C}\) is added to \(150.0 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\) at \(15.0^{\circ} \mathrm{C}\). The temperature of the water rises to \(18.3^{\circ} \mathrm{C}\). Calculate the specific heat capacity of the metal, assuming that all the heat lost by the metal is gained by the water.

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