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 61P
(a)
To determine
To Calculate: The angle of inclination of slide.
(b)
To determine
Velocity of the pig when it arrives at low end of the slide.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionStudents have asked these similar questions
A box is sliding with a constant speed of 4.00 m>s inthe +x-direction on a horizontal, frictionless surface. At x = 0 thebox encounters a rough patch of the surface, and then the surface becomeseven rougher. Between x = 0 and x = 2.00 m, the coefficientof kinetic friction between the box and the surface is 0.200; betweenx = 2.00 m and x = 4.00 m, it is 0.400. What is the x-coordinate of the point where the box comes to rest?
In a pickup game of dorm shuffleboard, students crazed by final exams use a broom to propel a calculus book along the dorm hallway. If the 3.5 kg book is pushed from rest through a distance of 0.92 m by the horizontal 26 N force from the broom and then has a speed of 1.28 m/s, what is the coefficient of kinetic friction between the book and floor?
A horizontal force of 12 newtons is applied to a 4.0 kg box that slides on a horizontal surface. The box starts from rest, moves a horizontal distance of 10 meters, and obtains a velocity of 5.0 m/s. The surface has friction. The friction force, assumed constant is what?
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 50.0kg object is pushed across a level surface with a diagonal force of 550 N causing it to move forward at a constant velocity of 6.00 m/s. Calculate the coefficient of kinetic friction. Can you please write it instead.arrow_forwardSome sliding rocks approach the base of a hill with a speed of12 m>s. The hill rises at 36° above the horizontal and has coefficientsof kinetic friction and static friction of 0.45 and 0.65, respectively, withthese rocks. Once a rock reaches its highest point, will it stay there or slide down the hill? If it stays, show why. If it slides, find its acceleration on the way down.arrow_forwardIf the coefficient of kinetic friction between tires and dry pavement is 0.80, what is the shortest distance in which you can stop a car by locking the brakes when the car is traveling at 28.7 m>s (about 65 mi>h)?arrow_forward
- Suppose that the coefficient of kinetic friction between Zak's feet and the floor, while wearing socks, is 0.250. Knowing this, Zak decides to get a running start and then slide across the floor. If Zak's speed is 3.00 m/sm/s when he starts to slide, what distance ddd will he slide before stopping? Use 9.81 m/s2m/s2 for the acceleration due to gravity.arrow_forwardA 5 kg box sits at rest at the bottom of a ramp that is 8 m long and that is inclined at 30° above the horizontal. Thecoefficient of kinetic friction is 0.40 and the coefficient of static friction is 0.50. What constant force, appliedparallel to the surface of the ramp is required to push the box to the top of the ramp in a time of 4 s?arrow_forwardYou are helping a friend move. Consider two boxes which have masses in factors ofm; one of mass 6m and the other of mass 7m. The coefficient of static friction betweenthe two boxes is very large. The coefficient of static friction between the boxes and thefloor is 0.7 and the coefficient of kinetic friction is 0.1. You place the large box on top ofthe small box and slide them down a 1m high ramp out of the truck. The ramp isfrictionless because it is covered with wheels. The box slides down this ramp smoothlyonto the horizontal grass with coefficient of kinetic friction is 0.142857142857143. Howfar (in meters) will the boxes slide before they stop?arrow_forward
- A horizontal force of 200 N is used to push a 50.0-kg packing crate a distance of 6.00 m on a rough horizontal surface. If the crate moves at constant speed, the coefficient of kinetic friction between the crate and surface.arrow_forwardA 25.0 kg box of textbooks rests on a loading ramp thatmakes an angle a with the horizontal. The coefficient of kinetic frictionis 0.25, and the coefficient of static friction is 0.35. At this angle, how fast will the box be moving after it has slid 5.0 m along the loading ramp?arrow_forwardA worker pushes a 35.0-kg package on a horizontal roller-belt conveyor until it reaches a certain speed. The package then coasts on the horizontal conveyor, all the while slowing down, until it comes to a complete stop after moving 1.20 m. While the package is coasting on the conveyor, the opposing friction force averages 6.00 N. Use the WET to find the speed of the package after the worker pushes it and just as it begins to coast. 0.625 m/s 0.759 m/s 0.600 m/s 0.510 m/s 0.802 m/s 0.641 m/sarrow_forward
- a girl is sliding down a slide while being pulled by a horizontal rope with a tension of 30N.The coefficient of kinetic friction between the girl and the slide is 0.2 and she has a mass of 75kg. Solve for her acceleration down the ramp.arrow_forwardA parachutist whose mass is 80 kg drops from a helicopter hovering 1000 m above the ground and falls toward the ground under the influence of gravity. Assume that the force due to air resistance is proportional to the velocity of the parachutist, with the proportionality constant b, = 30 N-sec/m when the chute is closed and b, = 90 N-sec/m when the chute is open. If the chute does not open until the velocity of the parachutist reaches 20 m/sec, after how many seconds will the parachutist reach the ground? Assume that the acceleration due to gravity is 9.81 m/ sec.arrow_forwardas a fish jump vertically out of the water, assume that only two significant forces act on it: an upward force F exerted by the tail fin and the downward force due to gravity. a force chinook salmon has a length of 1.50 m and a mass of 48.5 kg. if the fish is moving upward at 3.00 m/s as its head first breaks the surface and has an upward speed of 5.80 m/s after two-thirds of its length has left the surface, assume constant acceleration and determine the following. a. the salmon's acceleration b. the magnitude of the force F during this intervalarrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Physics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage LearningPrinciples of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
Physics for Scientists and Engineers: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
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