/*! 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} Q97PP Energy of locomotion. On flat gr... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Energy of locomotion. On flat ground, a 70 - kg person requires about 300 W of metabolic power to walk at a steady pace of 5 -km/hr (1.4 m/s). Using the same metabolic power output, that person can bicycle over the same ground at 15 - km/h.

How many times greater is the kinetic energy of the person when biking than when walking? Ignore the mass of the bike. (a) 1.7; (b) 3; (c) 8; (d) 9

Short Answer

Expert verified

The correct answer is option (d).

Step by step solution

01

Identification of given data

The mass of the person is m = 70 kg.

The velocity of walking is vwalking=5km/h.

The velocity of cycling is vcycling=15km/h.

The metabolic power is P = 300 W.

02

Concept/Significance of kinetic energy

The expression of kinetic energy is given by,

K=12mv2

Here, m is the mass of the body, and v is velocity of the body.

03

Determine the number of times the kinetic energy of the person when biking than when walking

Take the ratio of kinetic energy of a person in cycling and walking.

KcyclingKcycling=12mv2cycling12mv2walking=v2cyclingv2walking=15km/h25km/h2=9

Simplify further.

Kcycling=9Kwalking

Therefore, the kinetic energy of a person in cycling is times greater than walking.

Thus, the options (a), (b), and (c) are incorrect.

Hence, the correct answer is option (d).

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

You decide to visit Santa Claus at the north pole to put in a good word about your splendid behavior throughout the year. While there, you notice that the elf Sneezy, when hanging from a rope, produces a tension of 395.0 Nin the rope. If Sneezy hangs from a similar rope while delivering presents at the earth’s equator, what will the tension in it be? (Recall that the earth is rotating about an axis through its north and south poles.) Consult Appendix F and start with a free-body diagram of Sneezy at the equator.

Planet Vulcan.Suppose that a planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average orbit radius of Mercury. What would be the orbital period of such a planet? (Such a planet was once postulated, in part to explain the precession of Mercury’s orbit. It was even given the name Vulcan, although we now have no evidence that it actually exists. Mercury’s precession has been explained by general relativity.)

In each case, find the x- and y-components of vectorA→: (a)A→=5.0i^-6.3j^; (b)A→=11.2j^-9.91i^; (c)A→=-15.0i^+22.4j^; (d)A→=5.0B→, whereB→=4i^-6j^.

A cargo ship travels from the Atlantic Ocean (salt water) to Lake Ontario (freshwater) via the St. Lawrence River. The ship rides several centimeters lower in the water in Lake Ontario than it did in the ocean. Explain

While driving in an exotic foreign land, you see a speed limit sign that reads 180,000 furlongs per fortnight. How many miles per hour is this? (One furlong is 1/8 mile, and a fortnight is 14 days. A furlong originally referred to the length of a plowed furrow.)

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.