PS4-2

# PS4-2 - U 1 2 = T 2 T 1 so that m 2 gx 1 2 kx 2 = 1 2 m 1 v...

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Problem Set 4: Problem 2. Problem: Ab locko fmass m 2 sits on top of a block of mass m 1 . The blocks rest on a spring, which is not connected to the lower block. The spring constant is k . The upper block is suddenly removed and the lower block begins moving upward. (a) Determine the maximum speed of the lower block, V max . Express your answer as a function of m 1 , m 2 , k and gravitational acceleration, g . (b) Compute V max for m 1 g =0 . 5 lb, m 2 g =0 . 75 lb and k =9 lb/ft. Solution: The first thing we must do is solve for the initial compression of the spring, x o .W emu s tdoth i s because the spring force is F = k ( x x o ) for the motion that follows. Since the spring initially supports both weights, clearly kx o =( m 1 + m 2 ) g = x o = ( m 1 + m 2 ) g k (a) The initial and final values of the kinetic energy, T 1 and T 2 , respectively, are T 1 =0 and T 2 = 1 2 m 1 v 2 Both the spring and gravity do work so that U 1 2 = 8 x 0 [ k ( x x o )] dx m 1 gx = w 1 2 kx 2 + kx o x We e e e x = x x =0 m 1 gx = 1 2 kx 2 + k ( m 1 + m 2 ) g k x m 1 gx = 1 2 kx 2 +( m 1 + m 2 ) gx m 1 gx Simplifying, we conclude that U 1 2 = m 2

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Unformatted text preview: U 1 2 = T 2 T 1 so that m 2 gx 1 2 kx 2 = 1 2 m 1 v 2 Therefore, we find v 2 = 2 m 2 g m 1 x k m 1 x 2 Now, when v achieves its maximum value, so does v 2 . Thus, we can determine the maximum value of v by working with v 2 . dv 2 dx = 2 m 2 g m 1 2 k m 1 x and d 2 v 2 dx 2 < So, v 2 and v reach their maximum values when dv 2 /dx = 0 , wherefore x = m 2 g k Hence, there follows v 2 max = 2 m 2 g m 1 m 2 g k k m 1 p m 2 g k Q 2 = 2 m 2 2 g 2 m 1 k m 2 2 g 2 m 1 k = m 2 2 g 2 m 1 k Thus, the maximum velocity is v max = m 2 g m 1 k (b) Using the given values of m 1 g = 0 . 5 lb, m 2 g = 0 . 75 lb and k = 9 lb/ft, we have v max = m 2 g m 1 gk/g = . 75 lb (0 . 5 lb)(9 lb / ft) / (32 . 174 ft / sec 2 ) = 2 . ft sec...
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## This note was uploaded on 10/12/2009 for the course AME 301 at USC.

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PS4-2 - U 1 2 = T 2 T 1 so that m 2 gx 1 2 kx 2 = 1 2 m 1 v...

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