Example 4.13 Connected ObjectsGoal Use both the general method and the system approach to solvea connected two-body problem involving gravity and friction.Problem A block with mass m, = 3.90 kg and a ball with mass m,%D7.70 kg are connected by a light string that passes over a frictionlesspulley, as shown in Figure 4.23a. The coefficient of kinetic frictionbetween the block and the surface is 0.300.(a) Find the acceleration of the two objects and the tension in thestring.m2(b) Check the answer for the acceleration by using the systemapproach.Figure 4.23a Two objects connectedby a light string that passes over africtionless pulley.yStrategy Connected objects are handled by applying Newton'ssecond law separately to each object. The free-body diagrams for theblock and the ball are shown in Figure 4.23b, with the +x-direction tothe right and the +y-direction upwards. The magnitude of theacceleration for both objects has the same value, |a,| = |a,| = a.The block with mass m, moves in the positive x-direction, and theball with mass m, moves in the negative y-direction, so a, = -aŋ2:Using Newton's second law, we can develop two equations involvingthe unknowns T and a that can be solved simultaneously. In part (b),treat the two masses as a single object, with the gravity force on theball increasing the combined object's speed and the friction force onthe block retarding it. The tension forces then become internal anddon't appear in the second law.m2TmogFigure 4.23b Free-body diagrams forthe objects.Solution Solution(a) Find the acceleration of the objects and thetension in the string.Write the components of Newton's second law forthe cube of mass m1.EF, = T- f = m,a1EF, = n – m,g = 0%3DThe equation for the y-component givesn = m,g.Substitute this value for n and f,equation for the x-component.T- HM,g = m1a (1)Mkn into theApply Newton's second law to the ball, recallingthat a, = -a1.EF, = -m,g + T = m,a, = -mza, (2)Subtract Equation (2) from Equation (1),eliminating T and leaving an equation that can besolved for a, (substitution can also be used).m2g – Hm19 = (m, + m2)a,m1 + m2Substitute the given values to obtain theacceleration.(7.70 kg) (9.80 )- (0.300) (3.90 kg) (9.80 )3.90 kg + 7.70 kgm/s2aSubstitute the value into Equation (1) to find theT =tension T.(b) Find the acceleration using the systemapproach, where the system consists of two blocks.Apply Newton's second law to the system and solvefor a. (Use m,, m,, µµ n, and g as appropriate.)(m, + m2)a = m2g – Hn = m29 – HkM19%3D

Given: m1 = 3.9 kg m2 = 7.7 kg coefficient of kinetic friction, μk = 0.3

Problem A block with mass m, 7.70 kg are connected by a light string that passes over a frictionless pulley, as shown in Figure 4.23a. The coefficient of kinetic friction between the block and the surface is 0.300. 3.90 kg and a ball with mass m2 %3D %3D

Problem A block with mass m, 7.70 kg are connected by a light string that passes over a frictionless pulley, as shown in Figure 4.23a. The coefficient of kinetic friction between the block and the surface is 0.300. 3.90 kg and a ball with mass m2 %3D %3D

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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