A $0.20 k g$ stone is held $1.3 m$ above the top edge of a water well and then dropped into it. The well has a depth of $5.0 m$ . Relative to the configuration with the stone at the top edge of the well, what is the gravitational potential energy of the stone–Earth system (a) before the stone is released and (b) when it reaches the bottom of the well? (c) What is the change in gravitational potential energy of the system from release to reaching the bottom of the well?

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Solution

Use $U = m g y$ , where $y$ is measured relative to a reference level. Here, we measure $y$ to be relative to the top edge of the well, where we take $y = 0$ .
(a) $y = 1.3 m : U = m g y = (0.20 k g) (9.80 m / s^{2}) (1.3 m) = + 2.5 J$

(b) $y = - 5.0 m : U = m g y = (0.20 k g) (9.80 m / s^{2}) (- 5.0 m) = - 9.8 J$

(c) $Δ U = U_{f} - U_{i} = (- 9.8 J) - (2.5 J) = - 12.3 = - 12 J$

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Use U=mgy, where y is measured relative to a reference level. Here, we measure y to be relative to the top edge of the well, where we take y=0.(a) y=1.3m:U=mgy=(0.20kg)(9.80m/s2)(1.3m)=+2.5J(b) y=–5.0m:U=mgy=(0.20kg)(9.80m/s2)(–5.0m)=−9.8J(c) ΔU=Uf−Ui=(−9.8J)−(2.5J)=−12.3=−12J