plugin-Exam2-soln

plugin-Exam2-soln - Problem 1. Solution: Because we define...

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Unformatted text preview: Problem 1. Solution: Because we define the positive direction to be downward, when the weight hangs freely the gravitational force balances the spring force. Hence, mg k f = 0 = f = mg k (a) The velocity of the weight at both the start and the end of its motion is zero. Hence, the change in kinetic energy is zero. The work done by gravity in moving a distance z downward is U g = mgz . The work done by the spring is U s = 1 2 kz 2 . Thus, the Principle of Work and Energy tells us that mgz m 1 2 kz 2 m = 0 = z m = 2 mg k = 2 f (b) The spring force on the weight at maximum deflection is F s = kz m = 2 k mg k = 2 mg (c) Since only conservative forces are acting, we use conservation of mechanical energy. At maximum deflection, the weight is at rest so that its kinetic energy is zero. Also, its potential energy is 1 2 kz 2 m mgz m . Thus, as it moves upward, energy conservation tells us that 1 2 mv 2 + 1 2 kz 2 mgz = 0 + 1 2 kz 2 m mgz m = v 2 = k m p z 2 m z 2 Q + 2 g ( z z m ) Noting that k/m = g/ f and z m = 2 f , the velocity satisfies the following equation. v 2 = g f p 4 f 2 z 2 Q + 2 g ( z 2 f ) = 2 gz g z 2 f When v is a maximum, so is v 2 . Differentiation yields dv 2 dz = 2 g 2 g z f and d 2 v 2 dz 2 = 2 g f Hence, when dv 2 /dz = 0 , the second derivative is negative so that we have a maximum value of v 2 ....
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This note was uploaded on 05/11/2011 for the course AME 301 taught by Professor Shiflett during the Spring '06 term at USC.

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plugin-Exam2-soln - Problem 1. Solution: Because we define...

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