Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 5, Problem 2P
To determine
To Compare: The minimum acceleration at which the small block slips and at which heavier block slips.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionStudents have asked these similar questions
A man steps outside one winter day to go to work. His icydriveway is 8.0 m long from top to mailbox, and it slopes downwardat 20° from the horizontal. He sets his briefcase on theice at the top while opening the garage, and the briefcase slidesdown the driveway. Ignore friction. (a) What is its acceleration?(b) How many seconds does it take to get halfway to the mailbox?(c) How many seconds does it take to reach the mailbox?(d) What is its speed at the instant it reaches the mailbox?
The acceleration due to gravity on Earth differs slightly from place to place, depending on topography (e.g., whether you areon a hill or in a valley) and subsurface geology (whether there is dense rock like iron ore as opposed to light rock like saltbeneath you.) The precise acceleration due to gravity can be calculated from data taken in an introductory physicslaboratory course. An object, usually a metal ball for which air resistance is negligible, is dropped and the time it takes to falla known distance is measured. Very precise results can be produced with this method ifsufficient care is taken in measuring the distance fallen and the elapsed time.
A box of mass 100.0kg rests on level ground and there is no friction between the box and theground. Resting on top of that mass is a 40.0kg mass with friction coecient that's unknown. A ropeis tied to the 100.0 kg mass and draped over a pulley from which a third mass hangs. How much massmust the hanging mass have in order for the acceleration of all masses to be 5ms2 ?
What is the minimum possible friction coecient of the 40.0kg mass in the prior problem?
Chapter 5 Solutions
Physics for Scientists and Engineers
Ch. 5 - Prob. 1PCh. 5 - Prob. 2PCh. 5 - Prob. 3PCh. 5 - Prob. 4PCh. 5 - Prob. 5PCh. 5 - Prob. 6PCh. 5 - Prob. 7PCh. 5 - Prob. 8PCh. 5 - Prob. 9PCh. 5 - Prob. 10P
Ch. 5 - Prob. 11PCh. 5 - Prob. 12PCh. 5 - Prob. 13PCh. 5 - Prob. 14PCh. 5 - Prob. 15PCh. 5 - Prob. 16PCh. 5 - Prob. 17PCh. 5 - Prob. 18PCh. 5 - Prob. 19PCh. 5 - Prob. 20PCh. 5 - Prob. 21PCh. 5 - Prob. 22PCh. 5 - Prob. 23PCh. 5 - Prob. 24PCh. 5 - Prob. 25PCh. 5 - Prob. 26PCh. 5 - Prob. 27PCh. 5 - Prob. 28PCh. 5 - Prob. 29PCh. 5 - Prob. 30PCh. 5 - Prob. 31PCh. 5 - Prob. 32PCh. 5 - Prob. 33PCh. 5 - Prob. 34PCh. 5 - Prob. 35PCh. 5 - Prob. 36PCh. 5 - Prob. 37PCh. 5 - Prob. 38PCh. 5 - Prob. 39PCh. 5 - Prob. 40PCh. 5 - Prob. 41PCh. 5 - Prob. 42PCh. 5 - Prob. 43PCh. 5 - Prob. 44PCh. 5 - Prob. 45PCh. 5 - Prob. 46PCh. 5 - Prob. 47PCh. 5 - Prob. 48PCh. 5 - Prob. 49PCh. 5 - Prob. 50PCh. 5 - Prob. 51PCh. 5 - Prob. 52PCh. 5 - Prob. 53PCh. 5 - Prob. 54PCh. 5 - Prob. 55PCh. 5 - Prob. 56PCh. 5 - Prob. 57PCh. 5 - Prob. 58PCh. 5 - Prob. 59PCh. 5 - Prob. 60PCh. 5 - Prob. 61PCh. 5 - Prob. 62PCh. 5 - Prob. 63PCh. 5 - Prob. 65PCh. 5 - Prob. 67PCh. 5 - Prob. 68PCh. 5 - Prob. 69PCh. 5 - Prob. 70PCh. 5 - Prob. 71PCh. 5 - Prob. 72PCh. 5 - Prob. 73PCh. 5 - Prob. 74PCh. 5 - Prob. 75PCh. 5 - Prob. 76PCh. 5 - Prob. 77PCh. 5 - Prob. 78PCh. 5 - Prob. 79PCh. 5 - Prob. 80PCh. 5 - Prob. 82PCh. 5 - Prob. 83PCh. 5 - Prob. 84PCh. 5 - Prob. 85PCh. 5 - Prob. 86PCh. 5 - Prob. 87PCh. 5 - Prob. 88PCh. 5 - Prob. 89PCh. 5 - Prob. 90PCh. 5 - Prob. 91PCh. 5 - Prob. 92PCh. 5 - Prob. 93PCh. 5 - Prob. 94PCh. 5 - Prob. 95PCh. 5 - Prob. 96PCh. 5 - Prob. 97PCh. 5 - Prob. 101PCh. 5 - Prob. 102PCh. 5 - Prob. 103PCh. 5 - Prob. 104PCh. 5 - Prob. 105PCh. 5 - Prob. 106PCh. 5 - Prob. 107PCh. 5 - Prob. 108PCh. 5 - Prob. 109PCh. 5 - Prob. 110PCh. 5 - Prob. 111PCh. 5 - Prob. 112PCh. 5 - Prob. 113PCh. 5 - Prob. 114PCh. 5 - Prob. 115PCh. 5 - Prob. 116PCh. 5 - Prob. 117PCh. 5 - Prob. 118PCh. 5 - Prob. 119PCh. 5 - Prob. 120PCh. 5 - Prob. 121PCh. 5 - Prob. 122PCh. 5 - Prob. 123PCh. 5 - Prob. 124PCh. 5 - Prob. 125PCh. 5 - Prob. 126PCh. 5 - Prob. 127PCh. 5 - Prob. 128PCh. 5 - Prob. 129PCh. 5 - Prob. 130PCh. 5 - Prob. 131PCh. 5 - Prob. 132PCh. 5 - Prob. 133PCh. 5 - Prob. 134PCh. 5 - Prob. 135PCh. 5 - Prob. 136PCh. 5 - Prob. 137PCh. 5 - Prob. 138PCh. 5 - Prob. 139P
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.Similar questions
- A small package rest on the horizontal dashboard of a car. If us= 0.332, what the minimum acceleration of the car that will cause the package to slip off assuming that the is on a level read?arrow_forwardA box with mass m sits at the bottom of a long ramp thatis sloped upward at an angle a above the horizontal. You give the boxa quick shove, and after it leaves your hands it is moving up the rampwith an initial speed v0. The box travels a distance d up the ramp andthen slides back down. When it returns to its starting point, the speedof the box is half the speed it started with; it has speed v0 /2. What isthe coefficient of kinetic friction between the box and the ramp? (Youranswer should depend on only a.)arrow_forwardTwo boxes (Box A = 9.7 kg and Box B = 5.4 kg) are connected by acord running over a pulley. The coefficient of kinetic friction betweenbox A and the table is 0.18. We ignore the mass of the cord and pulleyand any friction in the pulley, which means we can assume that a forceapplied to one end of the cord will have the same magnitude at theother end. We wish to find the tension of the cord while accelerating (inN), assuming the cord doesn’t stretch. As box B moves down, box Amoves to the right. Hint: Solve for the acceleration of the system first.arrow_forward
- An elevator weighing 15kN starts from rest and acquires an upward velocity of 3m/sec in a distance of 6m. If the acceleration is constant, what is the tension in the cable?arrow_forwardThe terminal velocity of a human being is 200 km/hr, and it takes about 12 seconds to achieve this. This calculation considers for air friction of course. What would a person’s speed be if they free fell for 12 seconds instead (no air friction present)?arrow_forwardAn elevator in a tall building is allowed to reach a maximum speed of 3.5m/s going down. What must the tension be in the cable to stop this elevator over a distance of 2.6 m if the elevator has a mass of 1450 kg including occupants?arrow_forward
- A car weighs 1100kg and can accelerate from 0 to 60m/h in 7.1 seconds on a flat ground. what is the steepest grade your driveway can be if your house is located on a hill? Assume no friction forces, and approximate the driveway as a straight inclined plane.arrow_forwardA skier approaches the base of an icy hill with a speed of 12.5 m/s. The hill sloes upward at 24° above thehorizontal. Ignoring all friction forces, find the acceleration of this skier (a) when she is going up the hill (b) whenshe has reached her highest point (c) after she has started sliding down the hill.arrow_forwardYou are climbing on a hill that is inclined at 15degrees with the horizontal and you wish tobuild a rope tow to pull yourself up. Ignoring friction, calculate the tension you needed on therope to give your body that has a mass of 45kg a 1.5 m/s^2 acceleration. What if you are toconsider the friction of 100N, what is the value of the tension and coefficient of friction?arrow_forward
- A 51.5-kg swimmer with an initial speed of 1.25 m>s decides tocoast until she comes to rest. If she slows with constant acceleration and stops after coasting 2.20 m, what was the force exerted onher by the water?arrow_forwardSomeone at the other end of the table asks you to pass the salt. Feeling quite dashing, you slide the 50.0-g salt shaker inthat direction, giving it an initial speed of 1.15 m>s. (a) If the shaker comes to rest with constant acceleration in 0.840 m,what is the coefficient of kinetic friction between the shaker and the table? (b) How much time is required for the shakerto come to rest if you slide it with an initial speed of 1.32 m>s?arrow_forwardA 2.75 kg cat moves in astraight line (the x-axis). Figureshows a graph of the xcomponentof this cat’s velocityas a function of time. (a) Find themaximum net force on this cat.When does this force occur? (b)When is the net force on the catequal to zero? (c) What is the netforce at time 8.5 s?arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- College PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage LearningUniversity Physics (14th Edition)PhysicsISBN:9780133969290Author:Hugh D. Young, Roger A. FreedmanPublisher:PEARSONIntroduction To Quantum MechanicsPhysicsISBN:9781107189638Author:Griffiths, David J., Schroeter, Darrell F.Publisher:Cambridge University Press
- Physics for Scientists and EngineersPhysicsISBN:9781337553278Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningLecture- Tutorials for Introductory AstronomyPhysicsISBN:9780321820464Author:Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina BrissendenPublisher:Addison-WesleyCollege Physics: A Strategic Approach (4th Editio...PhysicsISBN:9780134609034Author:Randall D. Knight (Professor Emeritus), Brian Jones, Stuart FieldPublisher:PEARSON
College Physics
Physics
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
University Physics (14th Edition)
Physics
ISBN:9780133969290
Author:Hugh D. Young, Roger A. Freedman
Publisher:PEARSON
Introduction To Quantum Mechanics
Physics
ISBN:9781107189638
Author:Griffiths, David J., Schroeter, Darrell F.
Publisher:Cambridge University Press
Physics for Scientists and Engineers
Physics
ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:9780321820464
Author:Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:Addison-Wesley
College Physics: A Strategic Approach (4th Editio...
Physics
ISBN:9780134609034
Author:Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:PEARSON
Newton's Second Law of Motion: F = ma; Author: Professor Dave explains;https://www.youtube.com/watch?v=xzA6IBWUEDE;License: Standard YouTube License, CC-BY