
We can relate $\Delta E_{\text {th }}$ to the work $W$ done by $\vec{F}$ with the energy statement of Eq. 8-33 for a system that involves friction:
$$
W=\Delta E_{\text {mec }}+\Delta E_{\text {th }} .
$$
Calculations: We know the value of $W$ from (a). The change $\Delta E_{\text {mec }}$ in the crate's mechanical energy is just the change in its kinetic energy because no potential energy changes occur, so we have
$$
\Delta E_{\mathrm{mec}}=\Delta K=\frac{1}{2} m v^2-\frac{1}{2} m v_0^2 .
$$
Substituting this into Eq. 8-34 and solving for $\Delta E_{\mathrm{th}}$, we find
$$
\begin{aligned}
\Delta E_{\mathrm{th}} & =W-\left(\frac{1}{2} m v^2-\frac{1}{2} m v_0^2\right)=W-\frac{1}{2} m\left(v^2-v_0^2\right) \\
& =20 \mathrm{~J}-\frac{1}{2}(14 \mathrm{~kg})\left[(0.20 \mathrm{~m} / \mathrm{s})^2-(0.60 \mathrm{~m} / \mathrm{s})^2\right] \\
& =22.2 \mathrm{~J} \approx 22 \mathrm{~J} .
\end{aligned}
$$

A food shipper pushes a wood crate of cabbage heads (total mass $m=14 \mathrm{~kg}$ ) across a concrete floor with a constant horizontal force $\vec{F}$ of magnitude $40 \mathrm{~N}$. In a straight-line displacement of magnitude $d=0.50 \mathrm{~m}$, the speed of the crate decreases from $v_0=0.60 \mathrm{~m} / \mathrm{s}$ to $v=0.20 \mathrm{~m} / \mathrm{s}$. What is the increase $\Delta E_{\text {th }}$ in the thermal energy of the crate and floor?