Sm2 1thru2 3

PROBLEM 2.1\n\nKE = \\frac{1}{2} m v^{2}\\n\n = \\frac{1}{2} (0.316)(44)^{2}\\n = 27.581 \\, ft^{2}/s^{2}\\n\\nKE = 0.114 \\overline{B}h\\n\n\nPROBLEM 2.2\n\nKNOWN: An object of known mass is located at a specified elevation relative to the surface of the earth.\\nFIND: Determine gravitational potential energy of the object.\\n\\nE N G R E M O D E L: (1) The object is a closed system. (2) The acceleration of gravity is constant.\\n\\nm = 400 kg\\n\\ng = 9.78 m/s^{2}\\n\\nz = 25 m\\n\\nANALYSIS: The gravitational potential energy is\\n\\nPE = mg z\\n = (400 kg)(9.78 m/s^{2})(25 m)\\n = 97.8 kJ ⟵\\n\n\nPROBLEM 2.3\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n ΔKE = -500.1 lb\\nΔPE = +100.1 lb\\nv^{2} = 40ft/s\\nz = 30 ft\\n\\ng^{2} = mg = 102 lb/ft^{2}\\n\\n\n➀ ΔKE = \\frac{1}{2} m [V_{2}^{2} - V_{1}^{2}]\\n\\n\\n⟹ m =\\ frac{F_{avg}}{g} \\n\\n = \\frac{322.1 lb}{100 lb}\\n= 3.75 ft\\n\\n\n➁ ΔPE = mg(z_{2}-z_{1}) = 150.0 lb.ft = 100 lb(z_{2} - 30 ft)\\n\\nSolving,\\n z_{2} = 45 ft