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Engineering Mechanics - Dynamics Chapter 15 Problem 15-31 The log has mass M and rests on the ground for which the coefficients of static and kinetic friction are s and k respectively. The winch delivers a horizontal towing force T to its cable at A which varies as shown in the graph. Determine the speed of the log when t = t2. Originally the tension in the cable is zero. Hint: First determine the force needed to begin moving the log. Given: M = 500 kg t1 = 3 s s = 0.5 T1 = 1800 N k = 0.4 g = 9.81 m 2 s t2 = 5 s Solution: To begin motion we need t0 2 = Mg 2T 1 s t 2 1 t0 = s M g 2T 1 t1 t0 = 2.48 s Impulse - Momentum t1 0+ t 2 t 2T1 dt + 2T1 ( t2 t1 ) k M g( t2 t0 ) = M v2 t1 0 1 v2 = M t t1 t 2T1 dt + 2T1 ( t2 t1 ) k M g( t2 t0 ) t1 2 0 v2 = 7.65 m s *Problem 15-32 A railroad car having mass m1 is coasting with speed v1 on a horizontal track. At the same time another car having mass m2 is coasting with speed v2 in the opposite direction. If the cars meet and couple together, determine the speed of both cars just after the coupling. Find the difference between the total kinetic energy before and after coupling has occurred, and explain qualitatively what happened to this energy. 3 Units used: Mg = 10 kg Given: m1 = 15 Mg 3 kJ = 10 J m2 = 12 Mg 321 ... View Full Document
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