Answer
During the final $d=12 \mathrm{m}$ of motion, $W=0$ and we use
$$ K_{1}+U_{1} =K_{2}+U_{2}+f_{k} d $$
$$\frac{1}{2} m v^{2}+0 =0+0+f_{k} d $$
$$\text { where } m=0.42 \mathrm{kg} \text { and } v=4.2 \mathrm{m} / \mathrm{s} \text { gives } f_{k}=0.31 \mathrm{N} .$$
$$ \text { Therefore, the thermal energy change is} $$
$$ \Delta E_{\text {th }}=f_{k} d=3.7 \mathrm{J} $$
Work Step by Step
During the final $d=12 \mathrm{m}$ of motion, $W=0$ and we use
$$ K_{1}+U_{1} =K_{2}+U_{2}+f_{k} d $$
$$\frac{1}{2} m v^{2}+0 =0+0+f_{k} d $$
$$\text { where } m=0.42 \mathrm{kg} \text { and } v=4.2 \mathrm{m} / \mathrm{s} \text { gives } f_{k}=0.31 \mathrm{N} .$$
$$ \text { Therefore, the thermal energy change is} $$
$$ \Delta E_{\text {th }}=f_{k} d=3.7 \mathrm{J} $$
