/*! 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} Q97P A warehouse worker exerts a cons... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Q97P (page 148)

A warehouse worker exerts a constant horizontal force of magnitude 85 Non a 40 kgbox that is initially at rest on the horizontal floor of the warehouse. When the box has moved a distance of 1.4 m, its speed is 1.0 m/s. What is the coefficient of kinetic friction between the box and the floor?

Short Answer

Expert verified

渭=0.18

Step by step solution

01

Given                      

force=85Nonmass=40kgvi=0m/sandvf=1.0m/sx=1.4m

02

Understanding the concept

The problem deals with the frictional force. Frictional force is given by the product of coefficient of friction and the normal reaction. It also involves the kinematic equations of motion in which the motion of an object is described with constant acceleration.

Formula:

v2=v02+2ax鈭�x

03

Calculate the coefficient of kinetic friction between the box and the floor

F-fk=max,FN-mg=0

using v2=V02+2ax鈭�xwe find the acceleration be

ax=v2-v022鈭�x=1m/s2-021.4m=0.357m/s2

The above equations can be combined to give

fk=渭kFN渭k=fkFN=F-maxmg=85N-40kg0.357m/s240kg9.8m/s2=0.18

Thus,渭k=0.18

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 box of canned goods slides down a ramp from street level into the basement of a grocery store with acceleration0.75m/s2directed down the ramp. The ramp makes an angle of40掳with the horizontal. What is the coefficient of kinetic friction between the box and the ramp?

Block Ain Fig. 6-56 has mass mA=4.0kg, and block Bhas mass mB=2.0kg.The coefficient of kinetic friction between block B and the horizontal plane is 渭k=0.50.The inclined plane is frictionless and at angle 胃=30掳. The pulley serves only to change the direction of the cord connecting the blocks. The cord has negligible mass. Find

(a) the tension in the cord and

(b) the magnitude of the acceleration of the blocks.

Figure 6-16 shows the overhead view of the path of an amusement-park ride that travels at constant speed through five circular arcs of radii,R0,2R0and3R0. Rank the arcs according to the magnitude of the centripetal force on a rider traveling in the arcs, greatest first.

In the early afternoon, a car is parked on a street that runs down a steep hill, at an angle of35.0掳relative to the horizontal. Just then the coefficient of static friction between the tires and the street surface is 0.725. Later, after nightfall, a sleet storm hits the area, and the coefficient decreases due to both the ice and a chemical change in the road surface because of the temperature decrease. By what percentage must the coefficient decrease if the car is to be in danger of sliding down the street?

A car weighing 10.7kNand traveling at 13.4 m/swithout negative lift attempts to round an unbanked curve with a radius of 61.0 m. (a) What magnitude of the frictional force on the tires is required to keep the car on its circular path? (b) If the coefficient of static friction between the tires and the road is 0.350, is the attempt at taking the curve successful?
See all solutions

Recommended explanations on Physics Textbooks

Atoms and Radioactivity

Read Explanation

Particle Model of Matter

Read Explanation

Physics of Motion

Read Explanation

Circular Motion and Gravitation

Read Explanation

Electricity and Magnetism

Read Explanation

Electromagnetism

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.