429 Chapter 7 Conservation of Energy Conceptual Problems *1 • Determine the ConceptBecause the peg is frictionless, mechanical energy is conserved as this system evolves from one state to another. The system moves and so we know that ∆K> 0. Because ∆K+ ∆U= constant, ∆U< 0. correct.is)(a2 • Determine the ConceptChoose the zero of gravitational potential energy to be at ground level. The two stones have the same initial energy because they are thrown from the same height with the same initial speeds. Therefore, they will have the same total energy at all times during their fall. When they strike the ground, their gravitational potential energies will be zero and their kinetic energies will be equal. Thus, their speeds at impact will be equal. The stone that is thrown at an angle of 30°above the horizontal has a longer flight time due to its initial upward velocity and so they do not strike the ground at the same time.correct.is)(c3 • (a) False. Forces that are external to a system can do work on the system to change its energy. (b) False. In order for some object to do work, it must exert a force over some distance. The chemical energy stored in the muscles of your legs allows your muscles to do the work that launches you into the air. 4 • Determine the ConceptYour kinetic energy increases at the expense of chemical energy. *5• Determine the Concept As she starts pedaling, chemical energy inside her body is converted into kinetic energy as the bike picks up speed. As she rides it up the hill, chemical energy is converted into gravitational potential and thermal energy. While freewheeling down the hill, potential energy is converted to kinetic energy, and while braking to a stop, kinetic energy is converted into thermal energy (a more random form of kinetic energy) by the frictional forces acting on the bike.*6 • Determine the ConceptIf we define the system to include the falling body and the earth, then no work is done by an external agent and ∆K+ ∆Ug+ ∆Etherm= 0. Solving for the change in the gravitational potential energy we find ∆Ug= −(∆K+ friction energy).
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