# Chapter 8 - Potential Energy and Conservation of Energy - Problems: 13a

$0.98J$

#### Work Step by Step

Lets assume that the initial position of marble on the spring is A and that B is the point where the marble can go maximum vertically. The change in potential energy of the marble will be: $\Delta U_{g}=U_{B}-U_{A}=mgH_{B}-mgH_{A}=mg\Delta H$ From the question, we know that $\Delta H=20m$ ( marble travels maximum 20 m vertically). Therefore, $\Delta Ug=mg\Delta H\approx 5\times 10^{-3}kg\times 9.8\dfrac {m}{s^{2}}\times 20m\approx 0.98J$

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