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PROBLEM 2.4\n\nKNOWN: A brick of known volume and density experiences a given decrease in elevation.\n\nFIND: Determine the change in potential energy.\n\nSCHEMATIC & GIVEN DATA:\n\nENGR. MODEL: (1) The brick is a closed system. (2) The acceleration of gravity is constant. (3) The density of the body is uniform throughout.\n\nANALYSIS:\nV = (2.5 in)(3.5 in)(1 in) = 0.03038 ft³\n\nBased on assumption (3):\n\nm = 𝜌V = (120 lb/ft³)(0.03038 ft³) = 3.65 lb\n\nThe change in potential energy and the elevation are related by\nΔPE = mgΔz\n\n= (3.65 lb)(32.0 ft/s²)(-6.97 ft)\n\n= -25.0 ft·lb\n\nPROBLEM 2.5\n\nKNOWN: An automobile of known weight travels from sea level to a known elevation.\n\nFIND: Determine the change in potential energy.\n\nSCHEMATIC & GIVEN DATA:\n\nWeight = 2500 lb\n\nENGR. MODEL: 1. As shown in the schematic, the automobile is the system.\n2. The acceleration of gravity is constant.\n\nANALYSIS: The change in potential energy is\n\nΔPE = mg(z2 - z1)\n\nThe quantity mg is recognized as the vehicle's weight. Thus, inserting known values\n\nΔPE = (2500 lb)(7000 ft)(1 btu/778 ft·lb)\n\n= 2.25 × 10⁴ btu