## Fundamentals of Physics Extended (10th Edition)

Since only conservative forces are acting on the stone, we have that the mechanical energy of the system is conserved and thus $\Delta E_{mec} = \Delta U_{s} + \Delta U_{g} + \Delta K = 0$ where $\Delta U_{s}$ is the change in elastic potential energy, $\Delta U_{g}$ is the change in gravitational potential energy and $\Delta K$ is the change in kinetic energy. Since the stone is not moving before being released and also at the maximum height, we have that $\Delta K$ = 0. In Problem 19b, we found that $U_{s}$ before the stone was released was 62.7 J. At its maximum height, the stone is off the spring and all this energy has been converted to potential energy, so $U_{s}$ = 0. Therefore, $\Delta U_{s}$ = – 62.7 J. Substituting we get: $\Delta U_{s} + \Delta U_{g} + \Delta K = 0$ $–62.7 J + \Delta U_{g} + 0 = 0$ $\Delta U_{g}$ = 62.7 J So the change in the gravitational potential energy of the system is 62.7 J.