Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 8 - Potential Energy and Conservation of Energy - Problems - Page 203: 13b

Answer

– 0.98 J

Work Step by Step

Since only conservative forces are acting, we have that the mechanical energy $E_{mec}$ is conserved. Since kinetic, gravitational potential and elastic potential energies are involved, we have that: $\Delta E_{mec} = \Delta K + \Delta U_{g} + \Delta U_{s}$ = 0 From the moment the marble is fired to the moment it is at the top of its flight, the speed goes from 0 to 0 again, and thus $\Delta K$ = 0. In Problem 13a we found that $\Delta U_{s}$ = 0.98 J. Now we can just plug in these values and solve for the change in elastic potential energy $\Delta U_{s}$: 0 = $ \Delta K + \Delta U_{g} + \Delta U_{s}$ = 0 + 0.98 J +$ \Delta U_{s}$ $ \Delta U_{s}$ = –0.98 J
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