Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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 (2.0 J) than the object with a mass of 2.0 kg and a velocity of 1.0 m/s (1.0 J).

Step by step solution

01

Define the given parameters for each object

Object 1 has a mass (m1) of 2.0 kg and a velocity (v1) of 1.0 m/s. Object 2 has a mass (m2) of 1.0 kg and a velocity (v2) of 2.0 m/s.
02

Use the kinetic energy formula for each object

Calculate the kinetic energy (KE) of each object using the formula KE = (1/2)mv^2. For Object 1: KE1 = (1/2)m1v1^2 For Object 2: KE2 = (1/2)m2v2^2
03

Calculate the kinetic energies

Plug the given values of mass and velocity into the respective formulas and calculate the kinetic energies. For Object 1: KE1 = (1/2)(2.0 kg)(1.0 m/s)^2 KE1 = 1.0 J For Object 2: KE2 = (1/2)(1.0 kg)(2.0 m/s)^2 KE2 = 2.0 J
04

Compare the kinetic energies

Now that we have calculated the kinetic energies of both objects, it's time to compare them. From the calculations, KE1 is 1.0 Joule and KE2 is 2.0 Joules. As 2.0 J is greater than 1.0 J, we can conclude that the object with a mass of 1.0 kg and a velocity of 2.0 m/s (Object 2) has a greater kinetic energy than the object with a mass of 2.0 kg and a velocity of 1.0 m/s (Object 1).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Kinetic Energy Formula
The concept of kinetic energy is fundamental in physics, describing the energy an object possesses due to its motion. The formula for calculating this type of energy is a beautiful intersection of mass and velocity, expressed as \( KE = \frac{1}{2}mv^2 \). This equation reveals that kinetic energy (KE) is directly proportional to the mass (m) of the object and the square of its velocity (v). It's critical to understand that the velocity's influence on kinetic energy is exponential due to the squaring factor. This implies that even a small increase in velocity can result in a significant surge in kinetic energy.

Understanding the kinetic energy formula is crucial not just theoretically, but practically, as it explains phenomena ranging from the energy of a rolling ball to the destructive potential of a speeding bullet. It also demonstrates why high-speed collisions are much more hazardous than those at low speed; kinetic energy drastically increases as objects move faster, leading to more severe impacts.
Mass and Velocity
In the realm of kinetic energy, mass and velocity play pivotal roles. Mass is a measure of the amount of matter in an object and is often confused with weight, which is the force exerted by gravity on an object. Mass is intrinsic to an object and doesn't change unless the object itself changes in some material way. Velocity, however, is the speed of an object in a given direction, and it can change with the object's acceleration or deceleration.

Relationship Between Mass and Velocity in Kinetic Energy

When it comes to kinetic energy, mass provides the oomph' — more mass often means more energy. However, velocity is the real game-changer. Since kinetic energy depends on the velocity squared, even a lightweight object can surpass a more massive one in kinetic energy if it has a high enough velocity. This is exemplified by observing a bullet in motion versus a slow-moving boulder; despite their mass difference, the bullet can possess significantly greater kinetic energy due to its high velocity.
Energy Comparison
Comparing the kinetic energies of different objects can lead to some interesting and counterintuitive results, as seen in the exercise example. It reminds us that kinetic energy is not just about how heavy something is, but about how fast it's moving and how that speed scales in the computations.

Understanding Energy Comparisons

When comparing the kinetic energy of two objects using the kinetic energy formula, it might initially seem that the object with the greater mass will have more kinetic energy. However, we must consider that velocity has a squared effect in the equation. Therefore, when comparing the energy of objects with different masses and velocities, it's necessary to calculate the exact kinetic energy values before making a judgment. As demonstrated, an object with less mass but higher velocity can, in fact, have greater kinetic energy than a more massive, slower object. This understanding not only aids in solving physics problems but also enriches our perspective of how different physical factors interplay to determine how much energy an object in motion really has.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Explain the advantages and disadvantages of hydrogen as an alternative fuel.

Hydrogen gives off \(120 . \mathrm{J} / \mathrm{g}\) of energy when burned in oxygen, and methane gives off \(50 . \mathrm{J} / \mathrm{g}\) under the same circumstances. If a mixture of \(5.0 \mathrm{~g}\) hydrogen and \(10 . \mathrm{g}\) methane is burned, and the heat released is transferred to \(50.0 \mathrm{~g}\) water at \(25.0^{\circ} \mathrm{C}\), what final temperature will be reached by the water?

In a coffee-cup calorimeter, \(1.60 \mathrm{~g} \mathrm{NH}_{4} \mathrm{NO}_{3}\) is mixed with \(75.0 \mathrm{~g}\) water at an initial temperature of \(25.00^{\circ} \mathrm{C}\). After dissolution of the salt, the final temperature of the calorimeter contents is \(23.34^{\circ} \mathrm{C}\). Assuming the solution has a heat capacity of \(4.18 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) and assuming no heat loss to the calorimeter, calculate the enthalpy change for the dissolution of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) in units of \(\mathrm{kJ} / \mathrm{mol}\).

Combustion of table sugar produces \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\). When \(1.46 \mathrm{~g}\) table sugar is combusted in a constant-volume (bomb) calorimeter, \(24.00 \mathrm{~kJ}\) of heat is liberated. a. Assuming that table sugar is pure sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s)\), write the balanced equation for the combustion reaction. b. Calculate \(\Delta E\) in \(\mathrm{kJ} / \mathrm{mol} \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) for the combustion reaction of sucrose. c. Calculate \(\Delta H\) in \(\mathrm{kJ} / \mathrm{mol} \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) for the combustion reaction of sucrose at \(25^{\circ} \mathrm{C}\).

A biology experiment requires the preparation of a water bath at \(37.0^{\circ} \mathrm{C}\) (body temperature). The temperature of the cold tap water is \(22.0^{\circ} \mathrm{C}\), and the temperature of the hot tap water is \(55.0^{\circ} \mathrm{C}\). If a student starts with \(90.0 \mathrm{~g}\) cold water, what mass of hot water must be added to reach \(37.0^{\circ} \mathrm{C}\) ?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free