Answer
$$
\begin{aligned} \Delta E_{\text {th }} =0.53 \mathrm{J} \end{aligned}
$$
Work Step by Step
We use SI units so $m=0.075$ kg. Equation $8-33$ provides $\Delta E_{\text {th }}=-\Delta E_{\text {mec }}$ for the energy "lost" in the sense of this problem. Thus,
$$
\begin{aligned} \Delta E_{\text {th }} &=\frac{1}{2} m\left(v_{i}^{2}-v_{f}^{2}\right)+m g\left(y_{i}-y_{f}\right) \\ &=\frac{1}{2}(0.075 \mathrm{kg})\left[(12 \mathrm{m} / \mathrm{s})^{2}-(10.5 \mathrm{m} / \mathrm{s})^{2}\right]+(0.075 \mathrm{kg})\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right)(1.1 \mathrm{m}-2.1 \mathrm{m}) \\ &=0.53 \mathrm{J} \end{aligned}
$$
