SolHW3ME2320

# SolHW3ME2320 - 2 31 2 33 Given m = 100 kg mc = 100 kg = 20o...

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Unformatted text preview: 2 31 2 33. Given: m = 100 kg; mc = 100 kg; = 20o; g = 9.81 m/s2; x = 100 m Find: The work considering a) the man & b) the man and the cart. Solution: Assuming that the initial and final velocities are zero. The energy change in the system will correspond exclusively to the changes in potential energy. a) W= mgh = (100)(9.81)(100sin20) W = 33.552 kJ b) W=(m+mc)mgh= (100)(9.81)(100sin20) W = 67.104kJ 2 38. Given: ks = 3kN/cm; x = 3 cm. Find: W Solution: W=0.5k(x22 x12) = (.5)(300kN/m)(0.03m)2 W = 0.135 kJ ...
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## This note was uploaded on 04/07/2008 for the course ME 232 taught by Professor Monefort during the Spring '08 term at Western Michigan.

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SolHW3ME2320 - 2 31 2 33 Given m = 100 kg mc = 100 kg = 20o...

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