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EAS208_Exam02_102910_FINAL_Solutions

EAS208_Exam02_102910_FINAL_Solutions - EAS208 Fall 2010...

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EAS208 Fall 2010 – Exam 2 SOLUTIONS Name: Person Number: Problem 1: The collar is given a downward velocity of 2 kg 4 A m v s = when it is at . If the spring has an unstretched length of 1 and, and the velocity of the collar at is A m 1 s m = 5 C m v s = , determine the stiffness of the spring. k (30 points) Solution: The only forces that perform work on the collar are conservative. Apply Conservation of Total Mechanical Energy from position A to position C (assume the datum at position ): A 1 2 E E = gA eA A gC eC V V T V V T C + + = + + 2 2 2 1 1 1 1 0 2 2 2 2 A A C kS mv mgs kS mv + + = − + + 2 C ( ) ( ) 2 2 2 2 2 A C C A m v v mgs k S S + = ( ) ( ) 2 2 2 2 2 A C C A m v v mgs k S S + = where 4 A m v s = 5 C m v s = 2 2 1 A o S S = = = 1 m 2 2 2 2 2 2 1 1 C o S s S 5 1 m = + = + = 1 s m = Substitute the above expression for the spring stiffness relation: ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 4 5 2 2 9.81 1 40.24 5 1 1 A C C A m v v mgs N k k m S S + + = = = The stiffness of the spring is k 40.24 N k m = .
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EAS208 Fall 2010 – Exam 2 SOLUTIONS Name: Person Number: Problem 2: A pellet of mass 1.4 p m kg = is projected horizontally with a velocity 10 o m v s =
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