Part 4: Solve for a in terms of m, g, 0, and μkYOU WILL REQUIRE THE EQUATION EDITOR FOR THE GREEK SYMBOLS AND SUBSCRIPTS.Factorize as much as possible.a = g(sine-μcos) √Part 5: Does a depend on m?noCross-Check: Do the dimensions of your answer match that for acceleration?Discussion: Consider this problem with the x-axis parallel to the ground and y-axis perpendicular toit; the normal force N and force of friction f have both horizontal and vertical components. The sameis true with the acceleration. What happens if the coordinate system is rotated?Since coordinate axes are arbitrary, choose the one that simplifies the problem the most. Forinclines, this means that the x-axis is parallel to the incline, while the y-axis is perpendicular to it.Under this rotated coordinate system, which forces act purely in one direction? Which force has twonon-zero components? How can these be found by vector decomposition? Pay particular attention tothe angle of the incline and how that will affect the decomposition. [Fx=N-WyWx-for = ma (1)1. Drag and drop the heads and tails of the vectors to construct the free-body diagram.2. Note that angles do not need to be exact and magnitudes are not considered.=0 (2)Y1. An object of mass m is sliding down an incline that makes an angle with the horizontal with anacceleration a. The coefficient of kinetic friction between the block and the incline is Mk.Part 1: Set-up Newton's Second Law EquationsUsing the diagram in Part 1, set-up Newton's Second Law along the x and y axes in terms of thecomponents of the weight wxand Wy'the normal force N and the friction force f.What is the acceleration along the x direction? What is the acceleration along the y-direction?REMEMBER TO USE THE SUBSCRIPTS FOR W.REMEMBER TO USE THE CORRECT SIGNS ACCORDING TO THE COORDINATE SYSTEM IN PART 1.Xf; 153.43wy; 243.43°Wx; 333 43°ΣΕ, -Part 2: In terms or m, g and 0, the magnitude of the x and y components of weight w are given by:(i) Wx=

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Answered: Part 4: Solve for a in terms of m, g,…

Part 4: Solve for a in terms of m, g, 0, and μk. YOU WILL REQUIRE THE EQUATION EDITOR FOR THE GREEK SYMBOLS AND SUBSCRIPTS. Factorize as much as possible. a = g(sine-μcose)

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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esmat.h... Figure Lobby | Top Hat A uniform board of length L and mass M lies near a boundary that separates two regions. In region 1, the coefficient of kinetic friction between the board and the surface is μ, and in region 2, the coefficient is 2. The positive direction is shown in the figure. (Figure 1) Before: region 1 After: 141 Aktiv Chemistry FI 1₂ D2L Homepage - Vande... M Gmail 12 1 of 1 > region 2 4 Part A W = Submit Handshake VYES - Vanderbilt - S... Find the net work W done by friction in pulling the board directly from region 1 to region 2. Assume that the board me constant velocity. Express the net work in terms of M, g, L, ₁, and μ2. ►View Available Hint(s) IVE ΑΣΦ Part B Complete previous part(s) Provide Feedback Course Home P Pearson ? Review | Co Copyright ©2022 Pearson Education Inc. All rights reserved | Terms of Use | Privacy Policy Pormin ▬▬▬▬
A 3 kg block sits on an inclined plane. The angle of the plane is 30° from the ground. Because of static friction, the block is stationary. 1. Draw a free body diagram of the forces. 2. What is the weight of the block? 3. What is the normal force? 4. What is the force of friction? (Magnitude and direction).
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A block of mass M slides (3) along a rough (a) inclined surface (Ja au) with a constant velocity. The inclination angle ( ) of the surface is e as shown in the adjacent figure ). The coefficient (J) of kinetic friction (H ) of the Time left 059:40 inclined surface is: A. cos (e) B. tan(0) C. sin(20) M V- Constant D. cos(0) E. sin(e) *Take g= 9.8 m/s
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