F f v 1 = c100(10 3 ) m h da 1h 3600 s b =


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1 The 2Mg car has a velocity of v 1 = 100km>h when the v km/h driver sees an obstacle in front of the car. It takes 0.75 s for him to react and lock the brakes, causing the car to skid. If the car stops when it has traveled a distance of 175 m, determine the coefficient of kinetic friction between the tires and the road. Freeody Diagram: The normal reaction N on the car can be determined by writing the equation of motion along the y axis and referring to the freebody diagram of the car, Fig. a, + c F y = ma y ; N (9.81) = 2000(0) N = N Since the car skids, the frictional force acting on the car can be computed from F f = m k N = m k (19 620). Principle of Work and Energy: y referring to Fig. a, notice that only which is negative. The initial speed of the car is m>s. Here, the skidding distance of the car is s. T 1 + U 12 = T (2000)( ) + C m k (19 620)s D = 0 s = m k does work, The distance traveled by the car during the reaction time is s =v 1 t = 27.78(0.75) = m. Thus, the total tal distance traveled by the car before it stops is s = s +s 175 = m k F f v 1 = c100(10 3 ) m h da 1h 3600 s b = This work uriprotected s by United States copyright laws sale of any part of his work (including on the World Wide Web) m k = 0.255
2 The two blocks and have weights W = 60 lb and W = 10 lb. If the kinetic coefficient of friction between the incline and block is m k = 0.2, determine the speed of after it moves 3 ft down the plane starting from rest. Neglect the mass of the cord and pulleys. Kinematics: The speed of the block and can be related by using position coordinate equation s + (s  s ) = l 2s  s = l 2 s  s = 0 s = 2 s = 2(3) = 6ft 2v  v = 0 (1) Equation of Motion: pplying Eq. 13 7, we have + F y = ma y ; N  60a 4 5 b = (0) N = 48.0 lb Principle of Work and Energy: y considering the whole system, W which acts in the direction of the displacement does positive work. W and the friction force F f = m k N = 0.2(48.0) = 9.60 lb does negative work since they act in the opposite po direction to that of displacement Here, W is being displaced vertically (downward) d) 3 and W is being displaced vertically (upward) s. Since blocks are 5 s and at rest initially, T 1 = 0. pplying Eq. 14 7, we have 0 + W 3 5 s  F f s  W s = 1 v 2 m m 2 v work (3)R (3)  10(6) = v v2 2 32provided 2 1 and Eqs. (1) and (2) yields T 1 + a U 12 = T = 60v v v 2 2 protected 2 (2) This w is by United States copyrigh laws 2and is ro solely for the use of instructors in teaching their courses assessing student learning. Dissemination or sale of any part of o this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. v = 3.52 ft>s v = ft s
3 The man at the window wishes to throw the 30kg sack on the ground. To do this he allows it to swing from rest at to point C, when he releases the cord at u = 30. Determine the speed at which it strikes the ground and the distance R. 8m 8m 16 m u C T + U  C = T C (9.81)8 cos 30 = 1 2 (30)v2 C y C = m>s R D T + U  D = T D (9.81)(16) = 1 2 (30) v2 D v D = 17.7 m>s During free flight: (+T) s = s 0 + v 0 t a ct 2 16 = 8 cos sin 30 t (9.81)t2 t t = 0 Solving for the positive root: t = s ( : + ) s = s 0 + v 0 t s = 8 sin cos 30 (2.0784) s = m Thus, This work is protected by United States copyright laws sale of any part of this work (including on the World Wide Web) R = = 33.0 m lso, (v D ) x = cos 30 = m>s (+T)(v D ) x = sin (2.0784) = m>s n D = 2(10.097) 2 + (14.559) 2 = 7.7 m>s
4 lb block rests on the smooth semicylindrical surface. n elastic cord having a stiffness k = 2lb>ft is attached to the block at and to the base of the semicylinder at point C.If the block is released from rest at ( u = 0 ), determine the unstretched length of the cord so that the block begins to leave the semicylinder at the instant u = 45. Neglect the size of the block. C k 2lb/ft u 1.5 ft +b F n = ma n ; 2 sin 45 = a v2 1.5 b v = ft>s T 1 + U 12 = T (2)Cp(1.5)  l 0D (2)c 3p 2 4 (1.5)  l 0 d  2(1.5 sin 45 ) = 1 2 a b(5.844)2 l 0 = 2.77 ft This work is protected by United States copyright lawsws sale of any part of this work (including on the World Wide Web)
5 * The 500kg elevator starts from rest and travels upward with a constant acceleration a c = 2m>s 2. Determine the power output of the motor M when t = 3s. Neglect the mass of the pulleys and cable. M + c F y = ma y ; 3T  500(9.81) = 500(2) T = N E 3s E  s P = l 3 v E = v P When t = 3s, (+ c) v 0 + a c t v E = 0 + 2(3) = 6m>s v P = 3(6) = 18 m>s P O = (18) P O = 35.4 kw This work is protected by United States copyright laws sale of any part of this work (including on theworld Wide Web)
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