A Block-Spring SystemA block of mass 1.0 kg is attached to a horizontal spring that has a force constant of 2,000 N/m as shownfigure (a). The spring is compressed 3.0 cm and is then released from rest as in figure (b).(a) A block attached to a spring is pushed inwardfrom an initial position x-O by an externalagent. (b) At position x, the block is releasedfrom rest and the spring pushes it to the right.(For the following, when entering a mathematical expression, do not substitute numerical values; use variables only.)(a) Calculate the speed of the block as it passes through the equilibrium position x - 0 if the surface is frictionless.SOLUTIONConceptualize This situation has been discussed before, and it is easy to visualize the block being pushed to the right by the spring and moving with some speed at x = 0.Categorize We identify the system as the block and model the block as --Select-system.Analyze In this situation, the block starts with v, - 0 at x, = -3.0 cm, and we want to find v, at x, = 0.Use the following equation to find the work done by the spring with xay = x:2W maxWork is done on the block, and its speed changes. The conservation of energy equation reduces to the work-kinetic energy theorem. Use that theorem to find the speed at x = 0 (Use the followingnecessary: k, Xmay and m):W.Substitute numerical values to calculate the speed (in m/s):m/sFinalize Although this problem could have been solved in a previous chapter, it is presented here to provide contrast with the following part (b), which requires the techniques of this chapter.S-0-n- mg -0-n-Find the magnitude of the friction force (in N):k- Mn - Mng -INSubstitute 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 + AEm - Gmv? - 0) + 1,ax-fAx +Substitute numerical values (Enter your answer in m/s.):m/sFinalize As expected, this value is ---Select--- v 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 thisexample is equal to the increase in internal energy of the block-surface system in this example.Suppose the force Fis applied at an angle e 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 e-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--- the normal force, in turn reducing the force of friction, which suggests that the speed couldbe maximized by pulling at an angle other than e- 0.Categorize As in part (a), we model the block and the surface as a nonisolated system with a ---Select---v 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 e.):>Wother forces - WE- Fd cos(e) =(1)Apply the particle in equilibrium model to the block in the vertical direction: SE,-n+ Fsin(0) - mg = 0Solve for n (Use the following as necessary: F, Ax, 8, m and g.):(2)UseVother forces W = AK + AEnt to find the final kinetic energy for this situation:WF = AK + AEint = (Kp – 0) + fAx → Kp = WF - f;AxSubstitute the results in Equations (1) and (2):Ky = FAx cos(8) - HynAx = FAx cos(0) – Hg(mg - F sin(0))AxMaximizing 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.):dK,--FAx sin(0) – H(0 - F cos(8))Ax - 0d0-sin(8) + Hk cos(8) = 0- HkEvaluate e (in degrees) for H = 0.16:0 - tan(ug) = tan(0.16) =Finalize Notice that the angle at which the speed of the block is a maximum is indeed not e - 0. When the angle exceeds 9.09°, the horizontal component of the applied force is too small to be compensated by the reducedfriction force and the speed of the block begins to ---Select--- from its maximum value.EXERCISEA 17.5 kg block is dragged over a rough, horizontal surface by a 75 N force acting at 21° above the horizontal. The block is displaced 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.(s. Find the work done on the block bv the aravitational force.Jd) 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

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A Block-Spring System A block of mass 1.0 kg is attached to a horizontal spring that has a force constant of 2,000 N/m as shown in figure (a). The spring is compressed 3.0 cm and is then released from rest as in figure (b). (a) A block attached to a spring is pushed inward from an initial position x-O by an external agent. (b) At position x, the block is released from rest and the spring pushes it to the right. x=0

A Block-Spring System A block of mass 1.0 kg is attached to a horizontal spring that has a force constant of 2,000 N/m as shown in figure (a). The spring is compressed 3.0 cm and is then released from rest as in figure (b). (a) A block attached to a spring is pushed inward from an initial position x-O by an external agent. (b) At position x, the block is released from rest and the spring pushes it to the right. x=0

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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