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Chapter 6: Q.18 (page 155)

It takes the elevator in a skyscraper 4.0 s to reach its cruising speed of 10 m/s. A 60 kg passenger gets aboard on the ground floor. What is the passenger’s weight

a. Before the elevator starts moving?

b. While the elevator is speeding up?

c. After the elevator reaches its cruising speed?

Short Answer

Expert verified

a. Before the elevator starts moving N=588.6N

b. While the elevator is speeding up N=738.6N

c. After the elevator reaches its cruising speed N=588.6N

Step by step solution

01

Content Introduction

Inertia is a fundamental feature of all matter, and mass is a quantitative measure of it. The smaller the change caused by an applied force, the larger the mass of the body.

02

Explanation (Part a)

The elevator is at rest. The only two forces acting are gravitational and the force of the normal reaction of the substrate. N is the net force.

N=m×gN=60kg(9.81m/s2)N=588.6N

03

Explanation (Part b)

We will find the acceleration of the elevator from the equation written below:

v2=v1+atv1=0m/sv2=ata=v2ta=10m/s4sa=2.5m/s2N=m(a+g)N=60kg(9.81m/s2+2.5m/s2)N=738.6N

04

Explanation (Part c)

When the elevator reaches the maximum speed, it no longer accelerate.

ais acceleration ,Fis the force,Nis the net force.

a=0m/s2F=ma0=N-mgN=mgN=60kg(9.81m/s2)N=588.6N

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