/*! 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} Q.19 Zach, whose mass is 80 kg, is in... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Q.19 (page 155)

Zach, whose mass is 80 kg, is in an elevator descending at 10 m/s. The elevator takes 3.0 s to brake to a stop at the first floor.

a. What is Zach鈥檚 weight before the elevator starts braking?

b. What is Zach鈥檚 weight while the elevator is braking?

Short Answer

Expert verified

a. At rest fn=784N

b. Upon brakingrole="math" localid="1647497386095" fn=1050.4N

Step by step solution

01

Content Introduction

When the elevator is at rest or is moving at a constant speed, the net force that is acting on the Zach is zero. So that the apparent weight is equal to his real weight. Wappar=mg

02

Explanation (Part a)

When elevator is moving at a constant speed, Zach's apparent weight will be:

Wappar=m脳gWappar=80kg脳9.8m/s2Wappar=784N

03

Explanation (Part b)

When the elevator is going downward and is slowing down, the net force is upward. Notice that the apparent weight is normal force.

Wappar=ma+mg........(1)

Now we will find the value of acceleration. We are given:

v1=10m/svf=0m/st=3.0s

Therefore,

vf=vi+at0=10+(a脳3)a=-3.330m/s2

We will make it positive because its direction is upward.

From (1)

Wappar=mg+maWappar=m[g+a]Wappar=80脳(9.8+3.33)Wappar=1050.4N

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

Bonnie and Clyde are sliding a 300 kg bank safe across the floor to their getaway car. The safe slides with a constant speed if Clyde pushes from behind with 385 N of force while Bonnie pulls forward on a rope with 350 N of force. What is the safe鈥檚 coefficient of kinetic friction on the bank floor?
  1. Write a realistic problem for which these are the correct equations.
  2. Draw the free-body diagram and the pictorial representation for your problem.
  3. Finish the solution of the problem.
(100N)cos鈦�30鈭�鈭�fk=(20kg)ax

n+(100N)sin鈦�30鈭�鈭�(20kg)(9.80m/s2)=0fk=0.20n.

A 20kg loudspeaker is suspended 2.0 m below the ceiling by two 3.0-m-long cables that angle outward at equal angles. What is the tension in the cables?

A 8.0脳104kgspaceship is at rest in deep space. Its thrusters provide a force of 1200kN.The spaceship fires its thrusters for 20s, then coasts for12 km. How long does it take the spaceship to coast this distance?

A block of mass mis at rest at the origin at t=0. It is pushed with constant force F0from x=0to x=Lacross a horizontal surface whose coefficient of kinetic friction is 渭k=渭0(1鈭�x/L). That is, the coefficient of friction decreases from 渭0at x=0to zero at x=L.

a. Use what you've learned in calculus to prove that

ar=vrdvxdx

b. Find an expression for the block's speed as it reaches position L.
See all solutions

Recommended explanations on Physics Textbooks

Atoms and Radioactivity

Read Explanation

Thermodynamics

Read Explanation

Wave Optics

Read Explanation

Particle Model of Matter

Read Explanation

Oscillations

Read Explanation

Fluids

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.