/*! 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} Q12. When an object falls freely unde... [FREE SOLUTION] | 91影视

91影视

When an object falls freely under the influence of gravity, there is a net force mg exerted on it by the earth. Yet, by Newton鈥檚 third law, the object exerts an equal and opposite force on the earth. Does the earth move? Explain.

Short Answer

Expert verified

The earth does not move because of its larger mass as compared to the mass of the object.

Step by step solution

01

Step 1. Newton’s third law

When an object falls freely under the influence of gravity, there is a net force Fo=mgexerted on it by the earth. According to Newton鈥檚 third law, the object should also exert an equal and opposite force on the earth.

02

Step 2. Application of Newton’s second law

According to Newton鈥檚 second law, the expression for the force on the earth can be written as

FE=MaE.

Here, Mis the mass of the earth, and aEis the acceleration of the earth.

From the third law of motion,

FE=FoMaE=mgaE=mgM.

03

Step 3. Compare the masses of the earth and the object

As the mass of the earth is very large compared to the mass of the object, the acceleration of the earth becomes negligible, which is in the order of 10-25g. Hence, the object falls towards the earth, and the earth does not rise up towards the object.

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!

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

A bear sling (Fig. 4-40) is used in some national parks for placing backpackers鈥 food out of the reach of bears. As a backpacker raises the pack by pulling down on the rope, the force F needed

(a) decreases as the pack rises until the rope is straight across.

(b) doesn鈥檛 change.

(c) increases until the rope is straight.

(d) increases, but the rope always sags where the pack hangs.

A 75 kg petty thief wants to escape from a third-story jail window. Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg. How might the thief use this 鈥榬ope鈥 to escape? Give a quantitative answer.

Three blocks on a frictionless horizontal surface are in contact with each other, as shown in Fig. 4鈥54. A forceFis applied to block A (mass mA). (a) Draw a free-body diagram for each block. Determine (b) the acceleration of the system (in terms of mA, mB, and mC), (c) the net force on each block, and (d) the force of contact that each block exerts on its neighbor. (e) IfmA=mB=mC=10.0kg,andF=96.0N,give numerical answers for (b), (c), and (d). Explain how your answers make sense intuitively.

FIGURE 4-54 Problem 34

A stone hangs by a fine thread from the ceiling, and a section of the same thread dangles from the bottom of the stone (Fig. 4鈥36). If a person gives a sharp pull on the dangling thread, where is the thread likely to break: below the stone or above it? What if the person gives a slow and steady pull? Explain your answers.

FIGURE 4-36 Question 9

Arlene is to walk across a 鈥渉igh wire鈥 strung horizontally between two buildings 10.0 m apart. The sag in the rope when she is at the midpoint is 10.0掳, as shown in Fig. 4鈥47. If her mass is 50.0 kg, what is the tension in the rope at this point?

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.