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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’s weight before the elevator starts braking?

b. What is Zach’s 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×gWappar=80kg×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×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×(9.8+3.33)Wappar=1050.4N

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Most popular questions from this chapter

  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.

T−0.20n−(20kg)(9.80m/s2)sin⁡20∘

=(20kg)(2.0m/s2)n−(20kg)(9.80m/s2)cos⁡20∘=0

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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.

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