Answer
Using $f_{k}=0.31 \mathrm{N}$ for the entire distance $d_{\text {total }}=14 \mathrm{m},$ we obtain
$$
\Delta E_{\text {thotal }}=f_{k} d_{\text {total }}=(0.31 \mathrm{N})(14 \mathrm{m})=4.3 \mathrm{J}
$$
for the thermal energy generated by friction.
Work Step by Step
Using $f_{k}=0.31 \mathrm{N}$ for the entire distance $d_{\text {total }}=14 \mathrm{m},$ we obtain
$$
\Delta E_{\text {thotal }}=f_{k} d_{\text {total }}=(0.31 \mathrm{N})(14 \mathrm{m})=4.3 \mathrm{J}
$$
for the thermal energy generated by friction.
