What is the gravitational potential energy of a \(2.0-\mathrm{kg}\) book \(1.5 \mathrm{~m}\) above the floor?

Short Answer

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

Step by step solution

01

Identify the given values

In this problem, we are given the mass (m) of the book as 2.0 kg, the height (h) above the floor as 1.5 m, and the acceleration due to gravity (g) as approximately 9.81 m/s².
02

Write down the formula for gravitational potential energy

The formula to calculate gravitational potential energy is GPE = m * g * h
03

Plug in the given values into the formula

Now, we can substitute the values of m, g, and h into the formula: GPE = (2.0 kg) * (9.81 m/s²) * (1.5 m)
04

Multiply the values

Multiply the values together: GPE = 2.0 kg * 9.81 m/s² * 1.5 m = 29.43 kg m²/s²
05

Convert the gravitational potential energy to joules

Since 1 kg m²/s² is equivalent to 1 joule (J), the gravitational potential energy of the book is equal to 29.43 J. So, the gravitational potential energy of the 2.0 kg book 1.5 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!

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

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.)

A runner reaches the top of a hill with a speed of \(6.50 \mathrm{~m} / \mathrm{s}\) He descends \(50.0 \mathrm{~m}\) and then ascends \(28.0 \mathrm{~m}\) to the top of the next hill. His speed is now \(4.50 \mathrm{~m} / \mathrm{s}\). The runner has a mass of \(83.0 \mathrm{~kg} .\) The total distance that the runner covers is \(400 . \mathrm{m}\) and there is a constant resistance to motion of \(9.00 \mathrm{~N}\). Use energy considerations to find the work done by the runner over the total distance.

A father exerts a \(2.40 \cdot 10^{2} \mathrm{~N}\) force to pull a sled with his daughter on it (combined mass of \(85.0 \mathrm{~kg}\) ) across a horizontal surface. The rope with which he pulls the sled makes an angle of \(20.0^{\circ}\) with the horizontal. The coefficient of kinetic friction is \(0.200,\) and the sled moves a distance of \(8.00 \mathrm{~m}\). Find a) the work done by the father, b) the work done by the friction force, and c) the total work done by all the forces.

A snowboarder of mass \(70.1 \mathrm{~kg}\) (including gear and clothing), starting with a speed of \(5.1 \mathrm{~m} / \mathrm{s}\), slides down a slope at an angle \(\theta=37.1^{\circ}\) with the horizontal. The coefficient of kinetic friction is \(0.116 .\) What is the net work done on the snowboarder in the first 5.72 s of descent?

A cannonball of mass \(5.99 \mathrm{~kg}\) is shot from a cannon at an angle of \(50.21^{\circ}\) relative to the horizontal and with an initial speed of \(52.61 \mathrm{~m} / \mathrm{s}\). As the cannonball reaches the highest point of its trajectory, what is the gain in its potential energy relative to the point from which it was shot?

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