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Chapter 6: Problem 25

Relative to the ground, what is the gravitational potential energy of a 55.0 -kg person who is at the top of the Sears Tower, a height of \(443 \mathrm{~m}\) above the ground?

Short Answer

Expert verified
The gravitational potential energy is 238,379 J.

Step by step solution

01

Identify Known Values

In this problem, we are given some values. The mass of the person, \( m = 55.0 \text{ kg} \), and the height of the Sears Tower, \( h = 443 \text{ m} \). We can assume the gravitational acceleration \( g \) is \( 9.8 \text{ m/s}^2 \).
02

Understand the Formula for Gravitational Potential Energy

The gravitational potential energy \( U \) can be calculated using the formula: \[ U = mgh \] where \( U \) is the gravitational potential energy, \( m \) is the mass in kilograms, \( g \) is the gravitational acceleration, and \( h \) is the height in meters above the ground.
03

Substitute Known Values into the Formula

Replace the variables in the formula with the known values: \( m = 55.0 \text{ kg} \), \( g = 9.8 \text{ m/s}^2 \), and \( h = 443 \text{ m} \). This gives us: \( U = 55.0 \times 9.8 \times 443 \).
04

Perform the Calculation

Calculate the product: \[ U = 55.0 \times 9.8 \times 443 = 238,379 \text{ J} \] This gives us the gravitational potential energy in joules.

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Key Concepts

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

Physics Formula Derivation
Understanding the derivation of physics formulas is crucial for a deeper comprehension of concepts like gravitational potential energy. Gravitational potential energy is the energy possessed by an object due to its position relative to Earth. This is mathematically expressed with the formula \( U = mgh \). Let's break down each component of this equation:
  • \( U \) represents the gravitational potential energy. It is measured in joules (J).
  • \( m \) is the mass of the object in kilograms (kg). More mass means more potential energy for the same height.
  • \( g \) is the acceleration due to gravity, generally approximated as \( 9.8 \text{ m/s}^2 \) on Earth. This value can vary slightly depending on location.
  • \( h \) is the height above a reference point (often the ground) in meters (m). The higher the object, the greater the potential energy.
The formula \( U = mgh \) stems from the basic principle that gravitational force works over a distance, leading to potential energy accumulation as an object is lifted.
Potential Energy Calculation
Calculating potential energy involves using the derived formula to determine the energy stored by an object at a certain height. For the given exercise, we have a person with a mass of \( 55.0 \text{ kg} \) at a height of \( 443 \text{ m} \) above the ground. Using the gravitational potential energy formula \( U = mgh \), we can calculate:
  • Set \( m = 55.0 \text{ kg} \), \( g = 9.8 \text{ m/s}^2 \), and \( h = 443 \text{ m} \).
  • Plug these values into the formula: \( U = 55.0 \times 9.8 \times 443 \).
  • Calculate the result: \( U = 238,379 \text{ joules (J)} \).
This tells us how much energy is stored due to the person's height relative to the ground. It's important to remember that gravitational potential energy is a function of height, meaning it would change if the object's elevation changes.
Work-Energy Principle
The work-energy principle is a pivotal concept in understanding how energy transforms from one form to another. It states that work done on an object is equal to the change in its energy. In the context of gravitational potential energy, when you lift an object, you perform work on it against the force of gravity.Let's illustrate what happens:
  • When you lift a \( 55.0 \text{ kg} \) person to the height of the Sears Tower, you're doing work to overcome gravitational force, thereby increasing their potential energy.
  • The work done is significant because it directly translates into the gravitational potential energy stored in the person's elevated position.
  • If the person were to be lowered back to the ground, this potential energy would convert back to work done by gravity.
By understanding the work-energy principle, we see that energy isn't created or destroyed but simply converted from one form to another as work is done on or by objects. This highlights the conservation of energy law, which is foundational in physics.

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Most popular questions from this chapter

A swing is made from a rope that will tolerate a maximum tension of \(8.00 \times 10^{2} \mathrm{~N}\) without breaking. Initially, the swing hangs vertically. The swing is then pulled back at an angle of \(60.0^{\circ}\) with respect to the vertical and released from rest. What is the mass of the heaviest person who can ride the swing?

Multiple-Concept Example 5 reviews many of the concepts that play a role in this problem. An extreme skier, starting from rest, coasts down a mountain slope that makes an angle of \(25.0^{\circ}\) with the horizontal. The coefficient of kinetic friction between her skis and the snow is 0.200 . She coasts down a distance of \(10.4 \mathrm{~m}\) before coming to the edge of a cliff. Without slowing down, she skis off the cliff and lands downhill at a point whose vertical distance is \(3.50 \mathrm{~m}\) below the edge. How fast is she going just before she lands?

A 47.0 -g golf ball is driven from the tee with an initial speed of \(52.0 \mathrm{~m} / \mathrm{s}\) and rises to a height of \(24.6 \mathrm{~m}\). (a) Neglect air resistance and determine the kinetic energy of the ball at its highest point. (b)What is its speed when it is \(8.0 \mathrm{~m}\) below its highest point?

Two cars, \(A\) and \(B\), are traveling with the same speed of \(40.0 \mathrm{~m} / \mathrm{s}\), each having started from rest. Car A has a mass of \(1.20 \times 10^{3} \mathrm{~kg}\), and car \(\mathrm{B}\) has a mass of \(2.00 \times 10^{3} \mathrm{~kg} .\) Compared to the work required to bring car A up to speed, how much additional work is required to bring car B up to speed?

A helicopter, starting from rest, accelerates straight up from the roof of a hospital. The lifting force does work in raising the helicopter. (a) What type (s) of energy is (are) changing? Is each type increasing or decreasing? Why? (b) How is (are) the type(s) of energy that is (are) changing related to the work done by the lifting force? (c) If you want to determine the average power generated by the lifting force, what other variable besides the work must be known? An \(810-\mathrm{kg}\) helicopter rises from rest to a speed of \(7.0 \mathrm{~m} / \mathrm{s}\) in a time of \(3.5 \mathrm{~s}\). During this time it climbs to a height of \(8.2 \mathrm{~m}\). What is the average power generated by the lifting force?
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