/*! 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} Q102P The summit of Mount Everest is 8... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The summit of Mount Everest is 8850 mabove sea level. (a) How much energy would a 90 kgclimber expand against the gravitational force on him in climbing to the summit from sea level? (b) How many candy bars, at 1.25 MJper bar, would supply an energy equivalent to this? Your answer should suggest that work done against the gravitational force is a very small part of the energy expended in climbing a mountain.

Short Answer

Expert verified
  1. The energy that the climber expands to work against the gravitational force is7.8×106J
  2. Number of candy bars required are 6 bar .

Step by step solution

01

The given data

The mass of the climber is, m = 90 kg

The energy that the climber gets per bar, E = 1.25 MJ 106MJ1MJ=1.25×106J

The height of the summit of Mount Everest, h = 8850 m

The acceleration due to gravity is,g=9.8m/s2

02

Understanding the concept of energy

The energy of the climber against the gravitational force from the sea level shows the results that there is a change in potential energy.

Formula:

Change in potential energy, ∆PE=mgh2-h1 (1)

03

a) Calculation of the energy expended by the climber to climb the height

The work or energy expanded by the climber to go against the gravitational force to climb to the summit of the mountain is given using equation (1):

W = Change in potential energy

=90kg×9.8m/s2×8850m-0m=7.8×106kg.m2/s21J1kg.m2/s2=7.8×106J

Hence, the value of the energy is7.8×106J .

04

b) Calculation of the required candy bars to consume that energy in part (a)

The number of candy bars that are required by the climber to supply energy is given as:

n=7.8×106J1.25×106J/bar=6.2bar≈6bar

Hence, the required candies are 6 bar .

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 91Ó°ÊÓ!

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

When a click beetle is upside down on its back, it jumps upward by suddenly arching its back, transferring energy stored in a muscle to mechanical energy. This launching mechanism produces an audible click, giving the beetle its name. Videotape of a certain click-beetle jump shows that a beetle of mass m=4.0×10-6kgmoved directly upward by 0.77 mm during the launch and then to a maximum heighth = 0.30 m. During the launch, what are the average magnitudes of (a) the external force on the beetles back from the floor and (b) the acceleration of the beetle in terms of g?

A 70.0 kgman jumping from a window lands in an elevated fire rescue net 11.0 mbelow the window. He momentarily stops when he has stretched the net by 1.50 m. Assuming that mechanical energy is conserved during this process and that the net functions like an ideal spring, find the elastic potential energy of the net when it is stretched by 1.50 m.

During a rockslide, a 520 kgrock slides from rest down a hillside that islong and 300 mhigh. The coefficient of kinetic friction between the rock and the hill surface is 0.25. (a) If the gravitational potential energy Uof the rock–Earth system is zero at the bottom of the hill, what is the value of U just before the slide? (b) How much energy is transferred to thermal energy during the slide? (c) What is the kinetic energy of the rock as it reaches the bottom of the hill? (d) What is its speed then?

A 25 kg bear slides, from rest, 12 m down a lodge pole pine tree, moving with a speed of 5.6 m/s just before hitting the ground. (a) What change occurs in the gravitational potential energy of the bear-Earth system during the slide? (b) What is the kinetic energy of the bear just before hitting the ground? (c) What is the average frictional force that acts on the sliding bear?

In Fig. 8-21, a small, initially stationary block is released on a frictionless ramp at a height of 3.0 m. Hill heights along the ramp are as shown in the figure. The hills have identical circular tops, and the block does not fly off any hill. (a) Which hill is the first the block cannot cross? (b) What does the block do after failing to cross that hill? Of the hills that the block can cross, on which hill-top is (c) the centripetal acceleration of the block greatest and (d) the normal force on the block least?

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.