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
