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 = ½ (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) + ½ (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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 Spring '06
 Buchler
 Physics, Conservation Of Energy, Energy, Mass, Work, PEi, M1+ M2+, m1gh + m3g

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