Categorize The block is pulled by a force and the surface is rough, so the block and the surface aremodeled as ---Select--|system with a nonconservative force.A Block Pulled on a Rough SurfaceA 5.0 kg block initially at rest is pulled to the right along a horizontal surface by a constant horizontal force ofAnalyze Figure (a) illustrates this situation. Neither the normal force nor the gravitational force doeswork on the system because their points of application are displaced horizontally.21 N.Find the work done on the system by the applied force (Use the following as necessary: F and Ax.):(a) A block pulled to the right on a roughSWother forces- W.surface by a constant horizontal force.Apply the particle in equilibrium model to the block in the vertical direction (Use the following asnecessary: F, Ax, m, and g.):▼ mgEfy =0-n - mg = 0 → n =(b) The applied force is at an angle o to theFind the magnitude of the friction force (in N):horizontal.Substitute the energies into the equation shown and solve for the final speed of the block (Use thefollowing as necessary: F, Ax and f):FAx = AK + AEint - Gmv? - 0) + ,axmgV-1,Ax +m(For the following, when entering a mathematical expression, do not substitute numerical values; use variablesonly.)(a) Find the speed of the block after it has moved 2.5 m if the surfaces in contact have a coefficient of kineticfriction of 0.16.Substitute numerical values (Enter your answer in m/s.):]m/sSOLUTIONFinalize As expected, this value is ---Select--- v that found in the case of the block sliding on aConceptualizeThis example is similar to the example "A Block Pulled on a Frictionless Surface" in Chapter 7, butmodified so that the surface is no longer frictionless.frictionless surface. The difference in kinetic energies between the block sliding on a frictionless surfaceand the block in this example is equal to the increase in internal energy of the block-surface system inthis example.The rough surface applies a friction force on the block opposite to the applied force. As a result, weexpect the speed to be ---Select--- v that found in "A Block Pulled on a Frictionless Surface."(b) Suppose the force F is applied at an angle 0 as shown in figure (b). At what angle should the force beapplied to achieve the largest possible speed after the block has moved 2.5 m to the right? SOLUTIONConceptualize You might guess that 0 = 0 would give the largest speed because the force would havethe largest component possible in the direction parallel to the surface. Think about F applied at anarbitrary nonzero angle, however. Although the horizontal component of the force would be reduced, thevertical component of the force would ---Select--- v the normal force, in turn reducing the force offriction, which suggests that the speed could be maximized by pulling at an angle other than 8 = 0.Categorize As in part (a), we model the block and the surface as a nonisolated system with a---Select---0 = tan-(Hg) = tan-'(0.16) = |v force acting.Finalize Notice that the angle at which the speed of the block is a maximum is indeed not e = 0. Whenthe angle exceeds 9.09°, the horizontal component of the applied force is too small to be compensatedby the reduced friction force and the speed of the block begins to ---Select--- v from its maximumAnalyze Find the work done by the applied force, noting that d = Ax because the path followed by theblock is a straight line (Use the following as necessary: F, Ax, and 0.):value."other forces = W = Fd cos(0) =EXERCISE(1)A 17.5 kg block is dragged over a rough, horizontal surface by a 75 N force acting at 21° above the horizontal.Apply the particle in equilibrium model to the block in the vertical direction:The block is displaced 5.7 m, and the coefficient of kinetic friction is 0.200. (Enter your answers in J.)SFy =n + F sin(0) - mg = 0HintSolve for n (Use the following as necessary: F, Ax, 0, m and g.):(a) Find the work done on the block by the 75 N force.(2)n =(b) Find the work done on the block by the normal force.Use Wother forces - W = AK + AEnt to find the final kinetic energy for this situation:(c) Find the work done on the block by the gravitational force.WE = AK + AEint = (Kf - 0) + fAx → K = WF - fAxSubstitute the results in Equations (1) and (2):Kf = FAx cos(0) – HynAx = FAx cos(0) – Hg(mg – F sin(0))Ax(d) What is the increase in internal energy of the block-surface system due to friction?Maximizing the speed is equivalent to maximizing the final kinetic energy. Consequently, differentiate K,with respect to e and set the result equal to zero (Use the following as necessary: Ax, 0, and g.):dKf-= -FAx sin(0) - H(0 - F cos(0))Ax = 0de(e) Find the total change in the block's kinetic energy.-sin(0) + Hy cos(0) = 0= HNeed Help?Read ItEvaluate e (in degrees) for uk = 0.16:

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