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 of 21 N.(a) A block pulled to the right on a rough surface(b) The applied force is at an angle e to theby a constant horizontal force.horizontal.mgmg(For the following, when enteringmathematical expression, do not substitute numerical values; use variables only.)(a) Find the speed of the block after it has moved 2.5 m if the surfaces in contact have a coefficient of kinetic friction of 0.16.SOLUTIONConceptualizeThis example is similar to the example "A Block Pulled on a Frictionless Surface" in Chapter 7, but modified so that the surface is no longer frictionless.The rough surface appliesfriction force on the block opposite to the applied force. As a result, we expect the speed to be [higher thanthat found in "A Block Pulled on a Frictionless Surface."Categorize The block is pulled by a force and therough, so the blockd the surfacare modeled as a nonisated Vsystem with a nonconservative force.Analyze Figure (a) illustrates this situation. Neither the normal force nor the gravitational force does work on the system because their points of application are displaced horizontally.Find the work done on the system by the applied force (Use the following as necessary: F and Ax.):= W =other forcesApply the particle in equilibrium model to the block in the vertical direction (Use the following as necessary: F, Ax, m, and g.):>'F, = 0 -n - mg = 0 →n =Find the magnitudethe friction force (in N):f, = Hn = Hmg =NSubstitute the energies into the equation shown and solve for the final speed of the block (Use the following as necessary: F, Ax and f,.):FAx = AK + AE =Gmv? - 0) + fAxint-Ax +2V,=Substitute numerical values (Enter your answer in m/s.):m/sFinalize As expected, this value is -Select---that found in the case of the block sliding on a frictionless surface. The difference in kinetic energies between the block sliding on a frictionless surface and the block in this exampleequal to the increase in internal energy of the block-surface system in this example.(b) Suppose the force F is applied at an angle 0 as shown in figure (b). At what angle should the force be applied 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 have the largest component possible in the direction parallel to the surface. Think about Fapplied at an arbitrary nonzero angle, however. Although the horizontal component of the force would be reduced, the vertical component of the force would --Select--speed could be maximized by pulling at an angle other than 0 = 0.Categorize As in part (a), we model the block and the surface as a nonisolated system with a-Select---force acting.Analyze Find the work done by the applied force, noting that d = Ax because the path followed by the block is a straight line (Use the following as necessary: F, Ax, and 0.):(1) Swother forces W= Fd cos(0) =Apply the particle in equilibrium model to the block in the vertical direction: Apply the particle in equilibrium model to the block in the vertical direction:= n + F sin(0) - mg = 0Solve for n (Use the following as necessary: F, Ax, 0, m and g.):(2)n =Use= W = AK + AE,, to find the final kinetic energy for this situation:other forcesintW = AK + AE = (K,- 0) + f Ax - K,- W.-AxintSubstitute the results in Equations (1) and (2):K = FAx cos(0) - H,nAx = FAx cos(0) - u, (mg - F sin(0))AxMaximizing the speed is equivalent to maximizing the final kinetic energy. Consequently, differentiate K with respect to 0 and set the result equal to zero (Use the following as necessary: Ax, 0, and g.):dk,= -FAx sin(0) - H(0 - F cos(0))Ax = 0d0-sin(0) + H, cos(0) = 0Evaluate 0 (in degrees) for u, = 0.16:0 = tan-(H,) = tan-(0.16) =|Finalize Notice that the angle at which the speed of the block is a maximumindeed not 0 = 0. When the angle exceeds 9.09°, the horizontal component of the applied force is too small to be compensated by the reduced friction force and the speed of the block begins toSelect-from its maximum value.EXERCISEA 17.5 kg block is dragged overrough, horizontal surface by a 75 N force acting at 21° above the horizontal. The blockdisplaced 5.7 m, and the coefficient of kinetic friction is 0.200. (Enter your answers in J.)Hint(a) Find the work done on the block by the 75 N force.(b) Find the work done on the block by the normal force.(c) Find the work done on the block by the gravitational force.(d) What is the increase in internal energy of the block-surface system due to friction?(e) Find the total change in the block's kinetic energy.Need Help?Read It

Name: A Block Pulled on a Rough SurfaceA 5.0 kg block initially at rest is
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Description: 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 of 21 N.(a) A block pulled to the right on a rough surface(b) The applied force is at an angle e to th