Answer
$\text { We work this using English units (with } g=32 \mathrm{ft} / \mathrm{s})$ , but for consistency we convert the weight to pounds
$$
m g=(9.0) \mathrm{oz}\left(\frac{11 \mathrm{b}}{16 \mathrm{oz}}\right)=0.56 \mathrm{lb}
$$
which implies $m=0.018 \mathrm{lb} \cdot \mathrm{s}^{2} / \mathrm{ft}$ (which can be phrased as 0.018 $\mathrm{slug~as~explained~in~Appendix\ D)}$ . And we convert the initial speed to feet-per-second
$$
v_{i}=(81.8 \mathrm{mi} / \mathrm{h})\left|\frac{5280 \mathrm{ft} / \mathrm{mi}}{3600 \mathrm{s} / \mathrm{h}}\right|=120 \mathrm{ft} / \mathrm{s}
$$
or a more "direct" conversion from Appendix D can be used. Equation $8-30$ provides
$\Delta E_{\mathrm{th}}=-\Delta E_{\mathrm{mec}}$ for the energy "lost" in the sense of this problem. Thus,
$$\Delta E_{\mathrm{th}}=\frac{1}{2} m\left(v_{i}^{2}-v_{f}^{2}\right)+m g\left(y_{i}-y_{f}\right)$$$$=\frac{1}{2}(0.018)\left(120^{2}-110^{2}\right)+0$$$$=20 \mathrm{ft} \cdot \mathrm{lb}$$
Work Step by Step
$\text { We work this using English units (with } g=32 \mathrm{ft} / \mathrm{s})$ , but for consistency we convert the weight to pounds
$$
m g=(9.0) \mathrm{oz}\left(\frac{11 \mathrm{b}}{16 \mathrm{oz}}\right)=0.56 \mathrm{lb}
$$
which implies $m=0.018 \mathrm{lb} \cdot \mathrm{s}^{2} / \mathrm{ft}$ (which can be phrased as 0.018 $\mathrm{slug~as~explained~in~Appendix\ D)}$ . And we convert the initial speed to feet-per-second
$$
v_{i}=(81.8 \mathrm{mi} / \mathrm{h})\left|\frac{5280 \mathrm{ft} / \mathrm{mi}}{3600 \mathrm{s} / \mathrm{h}}\right|=120 \mathrm{ft} / \mathrm{s}
$$
or a more "direct" conversion from Appendix D can be used. Equation $8-30$ provides
$\Delta E_{\mathrm{th}}=-\Delta E_{\mathrm{mec}}$ for the energy "lost" in the sense of this problem. Thus,
$$\Delta E_{\mathrm{th}}=\frac{1}{2} m\left(v_{i}^{2}-v_{f}^{2}\right)+m g\left(y_{i}-y_{f}\right)$$$$=\frac{1}{2}(0.018)\left(120^{2}-110^{2}\right)+0$$$$=20 \mathrm{ft} \cdot \mathrm{lb}$$
