EAS208_HW05_101110_Solutions

EAS208_HW05_101110_Solutions - EAS 208(Fall 2010 Homework#5...

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EAS 208 (Fall 2010) Homework #5 – Solutions Problem 15–14 The force exerted on a car by a prototype crash barrier as the barrier crushes is , where ( 3 120 40 F s s = − + ) lb s is the distance in feet from the initial contact. The effective length of the barrier is 18 ft . How fast can a 5000 car be moving and be brought to rest within the effective length of the barrier? lb Answer: The maximum work down by the barrier on the car is given by: ( ) ( ) ( ) 18 18 2 4 3 2 4 1 2 0 0 120 40 60 10 60 18 10 18 1,069,200 ft ft U s s ds s s = + = = − = − ft lb Using the Work – Energy relation: 2 2 2 1 2 1 2 1 2 2 1 1 2 1 1 2 1 1 1 0 2 2 2 U U T U mv mv U mv v m = Δ = = = 1 2 1 1 1 2 2 2 1,069,200 117.35 5000 32.2 U ft lb ft v v v lb m s ft s × = = = or 1 80 mi v h = 1
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Problem 15–24 The system is released from rest. The mass slides on the smooth horizontal surface. By using the principle of work and energy, determine the magnitude of the velocity of the masses when the mass has fallen 1 . 4 kg 20 kg m Answer: Using the Work – Energy relation: 1 2 U T = Δ where 1 1 1 2 2 2 2 0 0 20 9.81 20 9.81 20 9.81 1 196.2 m m m m m m m U kg ds kg s kg m N s s s = × = × × = × × = m and ( ) ( ) ( ) ( ) 2 2 2 2 1 2 1 1 1 1 4 20 4 20 4 20 0 4 20 2 2 2 2 T v v v Δ = + + = + = + 2 2 v Thus: ( ) 2 1 2 2 2 2 1 2 196.2 196.2 4 20 4.04 2 24 m U T v v v s × = Δ = + = = Note that the tension of the rope is an internal force whose effect cancel outs and does not perform a net work on the system and that the velocities of the masses, while the system in motion, are the same. 2
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Problem 15–36 The ball is released from rest in position 1 with the string horizontal. The length of the string is . What is the magnitude of the ball’s velocity when it is in position ? 2 kg 1 L m = 2 Answer: Apply the Work – Energy relation for the motion of the ball from position 1 to a position where the string has an angle of 2 α with the horizontal: 1 2 U T = Δ where sin 1 2 2 0 sin 2 9.81 1 sin 40 12.61 L o m U mg ds mgL kg m N s α α = − = = × × × = j j m 3
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and 2 2 2 2 2 1 2 2 1 1 1 1 0 2 2 2 2 2 T mv mv mv v v Δ = = = = 2 2 Thus: 2 1 2 2 2 2 12.61 12.61 3.55 m U T v v v s = Δ = = = Note that the tension of the string does not perform any work on the ball because is always normal to the motion of the ball. The only force that performs any work on the ball is the gravitational force. Therefore, instead we could apply Conservation of Total Mechanical Energy (assume the datum at position 1 ): 1 2 E E = 1 1 2 g g V T V T 2 + = + 2 2 1 1 2 1 1 2 2 mgh mv mgh mv + = + 2 ( ) 2 2 1 0 0 sin 2 mg L mv α + = + 2 2 1 sin 2 mv mgL α = 2 2 sin v gL α = 2 2 9.81 1 sin 40 o v = × × × 2 3.55 m v s = Note that the same result as before is obtained. 4
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Problem 15–48 A small pellet of mass starts from rest at position 1 and slides down the smooth surface of the cylinder to position 2 , where .
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