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Unformatted text preview: = 1 2 (2m)v 2 + 1 2 I ω 2 + 0 Rolling: v = ω r = m ω 2 r 2 + 1 2 (mR 2 ) ω 2 = r 2 + R 2 2 m ω 2 ⇒ ω = 2gH r 2 + R 2 2 10. Energy Conservation: K f + U sf = K o + U so + W fric (Let x = 0 for relaxed spring.) 0 + 1 2 kx 2 = 0 + 1 2 kd 2 f k (d + x) = 1 2 kd 2 µ k mg(d + x) (using f k = µ k N = µ k mg) Rearrange: 1 2 kx 2 + (µ k mg) x + µ k mgd  1 2 kd 2 = ⇒ 1 2 (100 N/m) x 2 + (0.40)(50 N) x + (0.40)(50 N)(0.50 m)  1 2 (100 N/m)(0.50 m) = (50 N/m) x 2 + (20 N) x  2.5 Nm = Solve for x =  20 ± (20) 2 4(50)(2.5) 2(50) =  0.50 m OR + 0.10 m 1st root gives original position, so choose 2nd root: x = + 0.10 m Distance from wall = L o + x = 1.00 m + 0.10 m = 1.10 m...
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 Spring '07
 LECLAIR,A
 mechanics, Conservation Of Energy, kg, m/s

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