ME 242 Selected Answers Chapter 4

ME 242 Selected Answers Chapter 4 - 4.1 4.1.2 4.1.4 4.1.6...

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Unformatted text preview: 4.1 4.1.2 4.1.4 4.1.6 4.1.8 4.1.10 4.1.12 4.1.14 4.1.16 4.1.18 4.1.20 4.1.22 4.1.24 4.1.26 Kinetic Energy v = -14.97 i m/s x = 2.12 m/s Work done by F = 300 lb ft = 0.035 d = 3.48 m F = 511 lb v = 0.463 i m/s h = 9.54108 m x = 0.050 m 0.127 kg m h = 0.01 m + m 0 = 0.0929 rad vC = 44.2 m/s (a) YES, (b) YES, (c) YES 4.3 4.3.2 4.3.4 4.3.6 4.3.8 4.3.10 4.3.12 4.3.14 4.3.16 4.3.18 4.3.20 4.3.22 4.3.24 4.3.26 Power and Efficiency cost= 0.25 cents P = 30 hp h = 213 m 39.4 minutes faster with motor. % increase = 700% v2 = 8.34 mph tA = 1.31 s, hpavg = 47.5 hp, tB = 1.91 s, hpavg = 12 hp = 0.545 v = 6.68 m/s 4.3% increase with no drag. P = (55.5 W/s)t P = (-11, 104 N)v Pin = 4.91102 W 4.2 4.2.2 4.2.4 4.2.6 4.2.8 4.2.10 4.2.12 4.2.14 4.2.16 4.2.18 4.2.20 4.2.22 4.2.26 4.2.28 4.2.30 Potential Energies Conservative Forces and = 140 x = 0.384 m k = 2.79103 N/m v = 9.92 m/s h = 1.97 m v = 6.94 m/s The mass reaches B v B = 0.172 m/s v B = -0.742 m/s v=0 The addition of the friction slows the mass by 50% The stunt can be successfully completed. v = 7.95 m/s (a): = vD , r2 (b): r2 = v (r r-D ) 2 k + m D2 - 2DL + 2rL - r2 , (c): rmax = 1.483 m, 2 k (d): r = - m (r - L) + (vD) r3 1 = 46.6 N = 232.5 lb |h3 - L| = 0.642 m k = 292 N/m x = 2.0 ft m g 1 2 A 2 mA + mB v + mA g L - k +mB g L - -2Ff L - m g A + d + 1k k 2 m g A -L = k min 2 m g A k 2 2 2 2 4.2.32 4.2.34 4.2.38 4.2.40 4.2.42 4.2.44 mA gLmin + mB g Lmin + d 4.2.46 + 1 k L - Lmin 2 v = 5.83 m/s 70 ...
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This note was uploaded on 09/19/2009 for the course ME 242 taught by Professor Kam during the Spring '06 term at Nevada.

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