Answer
$$
\begin{aligned} \Delta E_{\text {th }} &=1.1 \times 10^{4} \mathrm{J} \end{aligned}
$$
That the angle of $25^{\circ}$ is nowhere used in this calculation is indicative of the fact that energy is a scalar quantity.
Work Step by Step
Equation $8-33$ provides $\Delta E_{\mathrm{th}}=-\Delta E_{\text {mec }}$ for the energy "lost" in the sense of this problem. Thus,
$$
\begin{aligned} \Delta E_{\text {th }} &=\frac{1}{2} m\left(v_{i}^{2}-v_{f}^{2}\right)+m g\left(y_{i}-y_{f}\right) \\ &=\frac{1}{2}(60 \mathrm{kg})\left[(24 \mathrm{m} / \mathrm{s})^{2}-(22 \mathrm{m} / \mathrm{s})^{2}\right]+(60 \mathrm{kg})\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right)(14 \mathrm{m}) \\ &=1.1 \times 10^{4} \mathrm{J} \end{aligned}
$$
That the angle of $25^{\circ}$ is nowhere used in this calculation is indicative of the fact that
energy is a scalar quantity.
