Conceptual Physics (12th Edition)
12th Edition
ISBN: 9780321909107
Author: Paul G. Hewitt
Publisher: PEARSON
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Textbook Question
Chapter 5, Problem 66RCQ
Here the stone is sliding down a friction-free incline.
(a) Identify the forces that act on it, and draw appropriate force vectors, (b) Use the parallelogram rule to construct the resultant force on the stone (carefully showing that it has a direction parallel to the incline-the same direction as the stone’s acceleration).
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
In which of the following situations do the forces on the body sum to zero?
Show that the acceleration of any object down an incline where friction behaves simply (that is, where fk=μkN)fk=μkN) is a=g(sinθ−μkcosθ).a=g(sinθ−μkcosθ). Note that the acceleration is independent of mass and reduces to the expression found in the previous problem when friction becomes negligibly small (μk=0).(μk=0).
A crate of mass 50 kg slides on a horizontal floor under the action of an applied horizontal force of 200 N and against a friction force of 120 N. Calculate (a) the acceleration, and (b) the applied horizontal force required to give the same acceleration with twice the mass and half the friction.
Chapter 5 Solutions
Conceptual Physics (12th Edition)
Ch. 5 - When you push against a wall with your fingers,...Ch. 5 - A boxer can hit a heavy bag with great force. Why...Ch. 5 - How many forces are required for an interaction?Ch. 5 - Prob. 4RCQCh. 5 - Consider hitting a baseball with a bat. If we call...Ch. 5 - Prob. 6RCQCh. 5 - Prob. 7RCQCh. 5 - Prob. 8RCQCh. 5 - Prob. 9RCQCh. 5 - Earth pulls down on you with a gravitational force...
Ch. 5 - Prob. 11RCQCh. 5 - Prob. 12RCQCh. 5 - Prob. 13RCQCh. 5 - Prob. 14RCQCh. 5 - Prob. 15RCQCh. 5 - Prob. 16RCQCh. 5 - Prob. 17RCQCh. 5 - Prob. 18RCQCh. 5 - Prob. 19RCQCh. 5 - Prob. 20RCQCh. 5 - Prob. 21RCQCh. 5 - Prob. 22RCQCh. 5 - Prob. 23RCQCh. 5 - Prob. 24RCQCh. 5 - Prob. 25RCQCh. 5 - Prob. 26RCQCh. 5 - Prob. 27RCQCh. 5 - Prob. 28RCQCh. 5 - Prob. 29RCQCh. 5 - Prob. 30RCQCh. 5 - 31. A van exerts a force on trailers of different...Ch. 5 - a. The accelerations of the boxes
b. The...Ch. 5 - 33. Three identical pucks, A, B, and C, are...Ch. 5 - Prob. 34RCQCh. 5 - 35. The photo shows Steve Hewitt and daughter...Ch. 5 - Prob. 36RCQCh. 5 - Prob. 37RCQCh. 5 - Prob. 38RCQCh. 5 - Prob. 39RCQCh. 5 - Prob. 40RCQCh. 5 - Prob. 41RCQCh. 5 - Prob. 42RCQCh. 5 - Prob. 43RCQCh. 5 - Prob. 44RCQCh. 5 - Prob. 45RCQCh. 5 - Prob. 46RCQCh. 5 - Prob. 47RCQCh. 5 - Prob. 48RCQCh. 5 - Prob. 49RCQCh. 5 - Suppose that two carts, one twice as massive as...Ch. 5 - Prob. 51RCQCh. 5 - Prob. 52RCQCh. 5 - Prob. 53RCQCh. 5 - Prob. 54RCQCh. 5 - Prob. 55RCQCh. 5 - Prob. 56RCQCh. 5 - Prob. 57RCQCh. 5 - Prob. 58RCQCh. 5 - Prob. 59RCQCh. 5 - Prob. 60RCQCh. 5 - Prob. 61RCQCh. 5 - Prob. 62RCQCh. 5 - The same stone is being accelerated vertically...Ch. 5 - Prob. 64RCQCh. 5 - Prob. 65RCQCh. 5 - Here the stone is sliding down a friction-free...Ch. 5 - The stone is at rest, interacting with both the...Ch. 5 - Prob. 68RCQCh. 5 - Prob. 69RCQCh. 5 - Prob. 70RCQCh. 5 - Prob. 71RCQCh. 5 - Prob. 72RCQCh. 5 - Prob. 73RCQCh. 5 - Prob. 74RCQCh. 5 - Prob. 75RCQCh. 5 - Prob. 76RCQCh. 5 - Prob. 77RCQCh. 5 - Prob. 78RCQCh. 5 - Prob. 79RCQCh. 5 - The strong man can withstand the tension force...
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
- When you stand with bare feet in a wet bathtub, the grip feels fairly secure, and yet a catastrophic slip is quite possible. Explain this in terms of the two coefficients of friction.arrow_forwardA box with a mass of 5 kg accelerates its speed in a straight line, increasing it from 5 m/s to 8 m/s, due to the application of an external force acting for a duration of 2.0 s. Calculate the average strength of this force.arrow_forwardIf a block of mass m lies on a wedge of mass M, here all surfaces are smooth, and the block slips on the wedge, will the wedge have any acceleration? If yes, explain why and of what magnitude?arrow_forward
- Assuming the mass is the same, how does the acceleration of an object depend on the force exerted on the object?arrow_forwardIf an object is moving, is it possible for the net force acting on it to be zero?arrow_forwardSolve for the equation of acceleration and tension for the given connected bodies.arrow_forward
- A contestant in a winter sporting event pulls a 43 kg block of ice in the positive horizontal direction with a rope over his shoulders across a frozen lake as shown in the figure. Assume the coefficients of static and kinetic friction are µs=0.1 and µk=0.03. Calculate the minimum force F he must exert to get the block sliding in newtons? What is its acceleration in m/s2 once it starts to move, if that force is maintained?arrow_forwardTwo blocks connected by a string are pulled across a horizontal surface by a force applied to one of the blocks, as shown. The coefficient of kinetic friction between the blocks and the surface is 0.25. If each block has an acceleration of 2.0 m/s2to the right, what is the magnitude F of the applied force?arrow_forwardBlock A has a mass of 10 kg, and blocks B and c have masses of 5 kg each. Knowing that the blocks are initially at rest and that B moves through 3 m in 2 s, determine (a) the magnitude of the force P, (b) the tension in the cord AD. Neglect the masses of the pulleys and axle friction.arrow_forward
- Shown below is a 10.0 kg block being pushed by a horizontal force F of magnitude 221.0 N. The coefficient of kinetic friction between the two surfaces is 0.60. Find the acceleration of the block. (Enter the magnitude in m/s^2.)________ m/s^2arrow_forwardWhen the Moon is directly overhead at sunset, the force by Earth on the Moon, FEM , is essentially at 90° to the force by the Sun on the Moon, FSM , as shown below. Given that FEM = 1.98 × 1020 N and FSM = 4.36 × 1020 N, all other forces on the Moon are negligible, and the mass of the Moon is 7.35 × 1022 kg, determine the magnitude of the Moon’s acceleration.arrow_forwardA person is skiing in a horizontal snow surface with a net acceleration of 0.2 meters per second per second. If his weight is 72 kg and his coefficient of dynamic friction is 0.05, determine his normal force.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Newton's First Law of Motion: Mass and Inertia; Author: Professor Dave explains;https://www.youtube.com/watch?v=1XSyyjcEHo0;License: Standard YouTube License, CC-BY