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

What is the gravitational potential energy of a 2.00 -kg book \(1.50 \mathrm{~m}\) above the floor?

Short Answer

Expert verified
Answer: The gravitational potential energy of the book is 29.43 Joules.

Step by step solution

01

Identifying the Variables

We are given the following values: - Mass (m) = 2.00 kg - Height (h) = 1.50 m - Acceleration due to gravity (g) = 9.81 m/s²
02

Write Down the Formula for Gravitational Potential Energy

We will use the formula: PE = m * g * h
03

Plug in the Values and Calculate Potential Energy

Substitute the given values into the formula: PE = (2.00 kg) * (9.81 m/s²) * (1.50 m)
04

Solve for Gravitational Potential Energy

Carrying out the multiplication, we get: PE = 29.43 J (Joules) The gravitational potential energy of the 2.00-kg book 1.50 meters above the floor is 29.43 Joules.

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.

Physics Problem Solving
Understanding physics involves recognizing the problems and breaking them down into manageable parts. To solve a physics problem, especially in mechanics, we follow a sequence of steps to arrive at a solution.

When solving for gravitational potential energy, the first step involves identifying the known variables. In our example, the mass of the book (m), the height above the floor (h), and the constant acceleration due to gravity (g) are our known quantities. Next, we look for appropriate formulas that link these variables—in this case, the formula for gravitational potential energy, which is PE = m * g * h.

Once the relevant equation is chosen, we 'plug in' the values of the known variables. Finally, the last step is computation, where the values are multiplied to yield the gravitational potential energy, with the unit of joules representing energy. By breaking down the problem into these steps, students can tackle complex concepts through an organized approach, reducing errors and enhancing comprehension.
Mechanical Energy
Mechanical energy is the sum of kinetic energy, which is energy due to motion, and potential energy, which is energy stored due to the position or configuration of an object. In our context, gravitational potential energy is a form of mechanical energy.

Gravitational potential energy (PE) is the energy an object possesses because of its position in a gravitational field. The higher the object is above the reference point (usually the ground), the greater its gravitational potential energy. It is crucial because it has the ability to do work—which in a practical sense, means that when the book falls, that stored energy can be transformed into kinetic energy and perhaps exert a force when it hits the ground.

The concept of conservation of mechanical energy tells us that in the absence of air resistance or other non-conservative forces, the total mechanical energy of an object remains constant as it moves. If a student holds a book at a certain height, the book has a certain amount of gravitational potential energy. If dropped, it converts to kinetic energy just before it hits the ground. The understanding of these energy transformations is fundamental in the study of physics.
Gravitational Force
Gravitational force is one of the four fundamental forces in nature. It is an attractive force that exists between any two masses, drawing them toward each other. Every object with mass exerts a gravitational pull on every other mass. The greater the mass, the stronger the force.

The force of gravity on earth is what imparts an acceleration of 9.81 m/s² to all objects, regardless of their mass. This acceleration is why the formula for gravitational potential energy includes the 'g' variable. The acceleration due to gravity is crucial when calculating the force exerted by gravity on an object, which leads to its potential energy when at a height.

In our textbook example, the 2.00-kg book experiences a gravitational force that grants it potential energy simply by being 1.50 meters above the floor. If released, gravity will convert all that stored potential energy into motion, accelerating the book back to the floor. The interplay of gravitational force and energy is a cornerstone in the field of physics, and understanding it is essential for students as they explore the forces that govern our universe.

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

A spring has a spring constant of \(80 . \mathrm{N} / \mathrm{m}\). How much potential energy does it store when stretched by \(1.0 \mathrm{~cm} ?\) a) \(4.0 \cdot 10^{-3} \mathrm{~J}\) c) 80 J e) 0.8 J b) 0.40 J d) \(800 \mathrm{~J}\)

Can the kinetic energy of an object be negative? Can the potential energy of an object be negative?

An arrow is placed on a bow, the bowstring is pulled back, and the arrow is shot straight up into the air; the arrow then comes back down and sticks into the ground. Describe all of the changes in work and energy that occur.

A block of mass 5.0 kg slides without friction at a speed of \(8.0 \mathrm{~m} / \mathrm{s}\) on a horizontal table surface until it strikes and sticks to a horizontal spring (with spring constant of \(k=2000 . \mathrm{N} / \mathrm{m}\) and very small mass \(),\) which in turn is attached to a wall. How far is the spring compressed before the mass comes to rest? a) \(0.40 \mathrm{~m}\) c) \(0.30 \mathrm{~m}\) e) \(0.67 \mathrm{~m}\) b) \(0.54 \mathrm{~m}\) d) \(0.020 \mathrm{~m}\)

The greenskeepers of golf courses use a stimpmeter to determine how "fast" their greens are. A stimpmeter is a straight aluminum bar with a V-shaped groove on which a golf ball can roll. It is designed to release the golf ball once the angle of the bar with the ground reaches a value of \(\theta=20.0^{\circ} .\) The golf ball (mass \(=1.62 \mathrm{oz}=0.0459 \mathrm{~kg}\) ) rolls 30.0 in down the bar and then continues to roll along the green for several feet. This distance is called the "reading." The test is done on a level part of the green, and stimpmeter readings between 7 and \(12 \mathrm{ft}\) are considered acceptable. For a stimpmeter reading of \(11.1 \mathrm{ft}\), what is the coefficient of friction between the ball and the green? (The ball is rolling and not sliding, as we usually assume when considering friction, but this does not change the result in this case.)

See all solutions

Recommended explanations on Physics 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