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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 the greater kinetic energy, at \(2.0 J\), compared to the first object's kinetic energy of \(1.0 J\).

Step by step solution

01

Calculate the kinetic energy of the first object

We are given the mass (\(m_1\)) and velocity (\(v_1\)) of the first object: \(m_1 = 2.0 kg, v_1 = 1.0 m/s\) Now, we will use the kinetic energy formula to determine its kinetic energy (\(KE_1\)): \(KE_1 = \frac{1}{2}m_1v_1^2\) \(KE_1 = \frac{1}{2} (2.0 kg) (1.0 m/s)^2\) \(KE_1 = (1.0 kg) (1.0 m^2/s^2)\) \(KE_1 = 1.0 J\) (Joules)
02

Calculate the kinetic energy of the second object

We are given the mass (\(m_2\)) and velocity (\(v_2\)) of the second object: \(m_2 = 1.0 kg, v_2 = 2.0 m/s\) Now, we will use the kinetic energy formula to determine its kinetic energy (\(KE_2\)): \(KE_2 = \frac{1}{2}m_2v_2^2\) \(KE_2 = \frac{1}{2} (1.0 kg) (2.0 m/s)^2\) \(KE_2 = (0.5 kg) (4.0 m^2/s^2)\) \(KE_2 = 2.0 J\) (Joules)
03

Compare the kinetic energies

Now that we have calculated the kinetic energies of both objects, we can compare them: \(KE_1 = 1.0 J\) \(KE_2 = 2.0 J\) Since \(KE_2 > KE_1\), the second object, with a mass of \(1.0 kg\) and a velocity of \(2.0 m/s\), 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, often symbolized as KE, is the energy that an object possesses due to its motion. It's a fundamental concept not only in physics but also in various engineering and scientific applications. The kinetic energy of an object is given by the formula:
\[\begin{equation}KE = \frac{1}{2}mv^2\end{equation}\]
where:
  • \(m\) represents the mass of the object,
  • de\(v\) denotes the velocity of the object,
  • and the factor \(\frac{1}{2}\) is a constant.
Understanding this relationship allows you to predict how an object's kinetic energy changes as its velocity or mass varies.
Calculating Kinetic Energy
To calculate kinetic energy, you simply insert the relevant values into the kinetic energy formula. Take a mass and a velocity, square the velocity, multiply it by the mass and then by one-half.
\for example, for an object with a mass of \(2.0\,\text{kg}\) moving with a velocity of \(1.0\,\text{m/s}\), the calculation would be:
\[\begin{equation}KE = \frac{1}{2} \cdot 2.0\,\text{kg} \cdot (1.0\,\text{m/s})^2 = 1.0\,\text{J}\end{equation}\]
which results in a kinetic energy of \(1.0\,\text{J}\) (Joules). It's crucial to maintain consistency in the units used (mass in kilograms, velocity in meters per second), so that the answer is in joules, the standard unit for energy.
Comparing Kinetic Energies
In situations where you're comparing the kinetic energies of different objects, it’s the calculated Joules that provide a direct comparison. With our earlier examples:
  • Object 1 (\(2.0\,\text{kg}\), \(1.0\,\text{m/s}\)) has a kinetic energy of \(1.0\,\text{J}\).
  • Object 2 (\(1.0\,\text{kg}\), \(2.0\,\text{m/s}\)) has a kinetic energy of \(2.0\,\text{J}\).
Despite Object 1 having a greater mass, the higher velocity of Object 2 results in it having the greater kinetic energy. The example demonstrates that the kinetic energy is more sensitive to changes in velocity because it is a squared term in the formula, highlighting that velocity has a greater impact than mass on the kinetic energy of an object.
Mass and Velocity Relationship
The relationship between mass and velocity in the kinetic energy formula indicates a squared dependency on velocity and a linear dependency on mass. This means that if you double the velocity of an object, its kinetic energy will increase by a factor of four (since \(2^2 = 4\)), while doubling the mass will only double the kinetic energy.

Importance of Velocity

This relationship has critical implications in everyday situations and in scientific contexts. Lightweight objects might possess large amounts of kinetic energy if they are moving fast enough.

Conservation of Kinetic Energy

In a closed system where no external forces are at work, the concept of the conservation of kinetic energy becomes very useful. This principle forms the foundation of many calculations in physics, particularly those involving collisions and propulsion.

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Most popular questions from this chapter

Write reactions for which the enthalpy change will be a. \(\Delta H_{\mathrm{f}}^{\circ}\) for solid aluminum oxide. b. The standard enthalpy of combustion of liquid ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l) .\) c. The standard enthalpy of neutralization of sodium hydroxide solution by hydrochloric acid. d. \(\Delta H_{\mathrm{f}}^{\circ}\) for gaseous vinyl chloride, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}(g)\). e. The enthalpy of combustion of liquid benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}(l)\). f. The enthalpy of solution of solid ammonium bromide.

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 gas absorbs \(45 \mathrm{~kJ}\) of heat and does \(29 \mathrm{~kJ}\) of work. Calculate \(\Delta E\).

A system releases \(125 \mathrm{~kJ}\) of heat while \(104 \mathrm{~kJ}\) of work is done on it. Calculate \(\Delta E\).

The equation for the fermentation of glucose to alcohol and carbon dioxide is $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)+2 \mathrm{CO}_{2}(g) $$ The enthalpy change for the reaction is \(-67 \mathrm{~kJ}\). Is the reaction exothermic or endothermic? Is energy, in the form of heat, absorbed or evolved as the reaction occurs?

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