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Unformatted text preview: 1 after it moves up 2.0 m. m 2 m 1 m 3 Solution: Initially: PE i = 0 (take the initial positions of m 1 and m 3 to be y = 0) KE i = 0 Finally: PE f = m 1 gh + m 3 g(h) where h=2 m is the distance moved up (down) by m 1 (m 1 ) K E f = (m 1 + m 2 + m 3 ) v 2 Conservation of energy: PE i + KE i = PE f + KE f ! m 1 gh + m 3 g(h) + (m 1 + m 2 + m 3 ) v 2 = 0 Solve for v: v = ( ) 3 2 1 1 3 2 m m m gh m m + + = 3.3 m/s...
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This note was uploaded on 09/25/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.
 Spring '06
 Buchler
 Physics, Mass, Work

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