# From equation 8 3 avitational potential energy is ua

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Unformatted text preview: y = 9;8 N)(7.25 em) = 1.07 J and Ub {1.5x )(7.62 cm) = 1.12 J above the zero energy. Problem 12. A carbon monoxide molecule can be modeled as a carbon atom and an oxygen atom connected by a spring. If a displacement ofthe carbon by 1.6x10-12 m from. its equilibrium position relative to the oxygen increases the molecule's potential energy by 0.015 eV, what is the spring constant? = = = = Solution In this model, the zero of potential energy is at the equilibrium separation for the molecule, so the spring constant can be calculated from Equation 8-4 and the energy, k = 2Ujx2 = 2(0.015 eV)(1.6xlO-19 J/eV)+ (1.6xl0-12 01)2 = 1.88 kN/m. Problem \0(' .. i\~blem Problem 9 Solution. 13. How far would you have to stretch a spring of spring constant k = 1.4 kN 1m until it stored 210 J of energy? Solution Assuming one starts stretching from the unstretched position (x = 0), Equation 8~4 gives x J2U jk = ,AGO-kg hiker ascending 125Q-m-high Camel's Sliump mountain in Vermont has potential energy ':;2.4x 105 J j the zero of potential energy is taken \~.tthe mountain top. W...
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