/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 In one day, a \(75-\mathrm{kg}\)... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 1

In one day, a \(75-\mathrm{kg}\) mountain climber ascends from the 1500 -m level on a vertical cliff to the top at 2400 \(\mathrm{m}\) . The next day, she descends from the top to the base of the cliff, which is at an elevation of 1350 \(\mathrm{m}\) . What is her change in gravitational potential energy (a) on the first day and \((b)\) on the second day?

Short Answer

Expert verified
First day: 661,500 J; Second day: -772,200 J.

Step by step solution

01

Understand the Formula for Gravitational Potential Energy Change

The gravitational potential energy change \( \Delta U \) is given by the formula: \[ \Delta U = m \cdot g \cdot \Delta h \] where \( m \) is the mass of the climber, \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)), and \( \Delta h \) is the change in height.
02

Calculate Change in Elevation for Day 1

On the first day, the climber moves from 1500 m to 2400 m, therefore the change in elevation \( \Delta h_1 \) is \( 2400 \text{ m} - 1500 \text{ m} = 900 \text{ m} \).
03

Calculate Potential Energy Change for Day 1

Substitute the known values into the formula for Day 1: \[ \Delta U_1 = 75 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 900 \, \text{m} \] \( \Delta U_1 = 661500 \, \text{Joules} \).
04

Calculate Change in Elevation for Day 2

On the second day, the climber descends from 2400 m to 1350 m, so the change in elevation \( \Delta h_2 \) is \( 1350 \text{ m} - 2400 \text{ m} = -1050 \text{ m} \).
05

Calculate Potential Energy Change for Day 2

Substitute the known values into the formula for Day 2: \[ \Delta U_2 = 75 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times (-1050) \, \text{m} \] \( \Delta U_2 = -772200 \, \text{Joules} \).

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 91影视!

Key Concepts

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

Understanding Energy Conservation
The principle of energy conservation is a foundational concept in physics, asserting that energy cannot be created or destroyed but only transformed from one form to another. In the context of gravitational potential energy, the energy associated with an object's position relative to a gravitational field, this means any change in height directly affects the potential energy.

For our mountain climber: - When ascending, she gains gravitational potential energy. - When descending, she loses gravitational potential energy.
This principle is crucial in physics problem-solving as it allows us to predict how potential energy changes in different situations without needing other external information.
Basics of Vertical Motion
Vertical motion refers to the movement of an object up or down under the influence of gravity. For objects like our climber moving vertically, gravity is the main force at play.

In vertical motion:
  • The object's speed during ascent decreases as gravitational potential energy increases.
  • Conversely, in descent, speed increases as the potential energy decreases.
Mathematically, this vertical motion affects the change in potential energy, which we calculate using the formula: \[ \Delta U = m \cdot g \cdot \Delta h \]- Here, \( m \) is mass, \( g \) is gravitational acceleration (9.8 m/s虏 on Earth), and \( \Delta h \) is the change in height. This formula succinctly links the physical movement to the energetic consequences.
Tackling Physics Problem Solving
Physics problems might seem daunting, but breaking them into steps can simplify the process. Let's use the mountain climber exercise as an example.

First, identify what you're solving for鈥攊n this case, the change in gravitational potential energy over two days. Then follow these steps:
  • Use given information: The climber's mass and the heights involved.
  • Calculate change in height \( \Delta h \) for each day.
  • Substitute these values into the formula for potential energy change \( \Delta U = m \cdot g \cdot \Delta h \).
  • Ensure each step connects logically to the next, allowing you to track your calculations and verify results.
By systematically following such a process, physics problems can be approached with confidence and clarity.

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

An ideal spring of negligible mass is 12.00 \(\mathrm{cm}\) long when nothing is attached to it. When you hang a \(3.15-\mathrm{kg}\) weight from it, you measure its length to be 13.40 \(\mathrm{cm}\) . If you wanted to store 10.0 \(\mathrm{J}\) of potential energy in this spring, what would be its total length? Assume that it continues to obey Hooke's law.

Bungee Jump. A bungee cord is 30.0 \(\mathrm{m}\) long and, when stretched a distance \(x,\) it exerts a restoring force of magnitude \(k x\) Your father-in- law (mass 95.0 \(\mathrm{kg}\) ) stands on a platform 45.0 \(\mathrm{m}\) above the ground, and one end of the cord is tied securely to his ankle and the other end to the platform. You have promised him that when he steps off the platform he will fall a maximum distance of only 41.0 \(\mathrm{m}\) before the cord stops him. You had several bungee cords to select from, and you tested them by stretching them out, tying one end to a tree, and pulling on the other end with a force of 380.0 \(\mathrm{N}\) . When you do this, what distance will the bungee cord that you should select have stretched?

A crate of mass \(M\) starts from rest at the top of a frictionless ramp inclined at an angle \(\alpha\) above the horizontal. Find its speed at the bottom of the ramp, a distance \(d\) from where it started. Do this in two ways: (a) Take the level at which the potential energy is zero to be at the bottom of the ramp with \(y\) positive upward. (b) Take the zero level for potential energy to be at the top of the ramp with y positive upward. (c) Why did the normal force not enter into your solution?

A spring of negligible mass has force constant \(k=\) 1600 \(\mathrm{N} / \mathrm{m}\) . (a) How far must the spring be compressed for 3.20 \(\mathrm{J}\) of potential energy to be stored in it? (b) You place the spring vertically with one end on the floor. You then drop a \(1.20-\mathrm{kg}\) book onto it from a height of 0.80 \(\mathrm{m}\) above the top of the spring. Find the maximum distance the spring will be compressed.

A 72.0 -kg swimmer jumps into the old swimming hole from a diving board 3.25 \(\mathrm{m}\) above the water. Use energy conservation to find his speed just he hits the water (a) if he just holds his nose and drops in, (b) if he bravely jumps straight up (but just beyond the board) at 2.50 \(\mathrm{m} / \mathrm{s}\) , and (c) if he manages to jump downward at 2.50 \(\mathrm{m} / \mathrm{s} .\)
See all solutions

Recommended explanations on Physics Textbooks

Astrophysics

Read Explanation

Rotational Dynamics

Read Explanation

Oscillations

Read Explanation

Electromagnetism

Read Explanation

Physical Quantities And Units

Read Explanation

Waves Physics

Read Explanation
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.