Let x be the distance past P.

\(\displaystyle\mu_{{{b}}}={0.100}+{A}{x}\)

When \(\displaystyle{x}={12.5}{m}\mu_{{{b}}}={0.600}\)

\(\displaystyle{A}={\frac{{{0.500}}}{{{12.5}{m}}}}=\frac{{0.0400}}{{m}}\)

a)

\(\displaystyle{W}=\triangle{K}{E}:{W}_{{{f}}}={K}{E}_{{{f}}}-{K}{E}_{{{i}}}\)

\(\displaystyle-\int\mu_{{{b}}}{m}{g}{\left.{d}{x}\right.}={0}-{\frac{{{1}}}{{{2}}}}{m}{{v}_{{{i}}}^{{{2}}}}\)

\(\displaystyle{g}{\int_{{{0}}}^{{{x}_{{{f}}}}}}{\left({0.100}+{A}{x}\right)}{\left.{d}{x}\right.}={\frac{{{1}}}{{{2}}}}{{v}_{{{i}}}^{{{2}}}}\)

\(\displaystyle{g}{\left[{\left({0.100}\right)}{x}_{{{f}}}+{A}{\frac{{{{x}_{{{f}}}^{{{2}}}}}}{{{2}}}}\right]}={\frac{{{1}}}{{{2}}}}{{v}_{{{i}}}^{{{2}}}}\)

\(\displaystyle{\left({9.80}\frac{{m}}{{s}^{{{2}}}}\right)}{\left[{\left({0.100}\right)}{x}_{{{f}}}+{\left(\frac{{0.0400}}{{m}}\right)}{\frac{{{{x}_{{{f}}}^{{{2}}}}}}{{{2}}}}\right]}={\frac{{{1}}}{{{2}}}}{\left({4.50}\frac{{m}}{{s}}\right)}^{{{2}}}\)

Now solve for \(\displaystyle{x}_{{{f}}}\), We get the answer as \(\displaystyle{x}_{{{f}}}={5.11}{m}\)

b)

\(\displaystyle\mu_{{{k}}}={0.100}+{\left(-.-\frac{{40}}{{m}}\right)}{\left({5.11}{m}\right)}={0.304}\)

c)

\(\displaystyle{W}_{{{f}}}={K}{E}_{{{f}}}-{K}{E}_{{{i}}}\)

\(\displaystyle-\mu_{{{b}}}{m}{g}{x}={0}-{\frac{{{1}}}{{{2}}}}{m}{{v}_{{{1}}}^{{{2}}}}\)

\(\displaystyle{x}={\frac{{{{v}_{{{1}}}^{{{2}}}}}}{{{2}\mu_{{{b}}}{g}}}}={\frac{{{\left({4.50}\frac{{m}}{{s}}\right)}^{{{2}}}}}{{{2}{\left({0.100}\right)}{\left({9.8}\frac{{m}}{{s}^{{{2}}}}\right)}}}}={10.3}{m}\)

\(\displaystyle\mu_{{{b}}}={0.100}+{A}{x}\)

When \(\displaystyle{x}={12.5}{m}\mu_{{{b}}}={0.600}\)

\(\displaystyle{A}={\frac{{{0.500}}}{{{12.5}{m}}}}=\frac{{0.0400}}{{m}}\)

a)

\(\displaystyle{W}=\triangle{K}{E}:{W}_{{{f}}}={K}{E}_{{{f}}}-{K}{E}_{{{i}}}\)

\(\displaystyle-\int\mu_{{{b}}}{m}{g}{\left.{d}{x}\right.}={0}-{\frac{{{1}}}{{{2}}}}{m}{{v}_{{{i}}}^{{{2}}}}\)

\(\displaystyle{g}{\int_{{{0}}}^{{{x}_{{{f}}}}}}{\left({0.100}+{A}{x}\right)}{\left.{d}{x}\right.}={\frac{{{1}}}{{{2}}}}{{v}_{{{i}}}^{{{2}}}}\)

\(\displaystyle{g}{\left[{\left({0.100}\right)}{x}_{{{f}}}+{A}{\frac{{{{x}_{{{f}}}^{{{2}}}}}}{{{2}}}}\right]}={\frac{{{1}}}{{{2}}}}{{v}_{{{i}}}^{{{2}}}}\)

\(\displaystyle{\left({9.80}\frac{{m}}{{s}^{{{2}}}}\right)}{\left[{\left({0.100}\right)}{x}_{{{f}}}+{\left(\frac{{0.0400}}{{m}}\right)}{\frac{{{{x}_{{{f}}}^{{{2}}}}}}{{{2}}}}\right]}={\frac{{{1}}}{{{2}}}}{\left({4.50}\frac{{m}}{{s}}\right)}^{{{2}}}\)

Now solve for \(\displaystyle{x}_{{{f}}}\), We get the answer as \(\displaystyle{x}_{{{f}}}={5.11}{m}\)

b)

\(\displaystyle\mu_{{{k}}}={0.100}+{\left(-.-\frac{{40}}{{m}}\right)}{\left({5.11}{m}\right)}={0.304}\)

c)

\(\displaystyle{W}_{{{f}}}={K}{E}_{{{f}}}-{K}{E}_{{{i}}}\)

\(\displaystyle-\mu_{{{b}}}{m}{g}{x}={0}-{\frac{{{1}}}{{{2}}}}{m}{{v}_{{{1}}}^{{{2}}}}\)

\(\displaystyle{x}={\frac{{{{v}_{{{1}}}^{{{2}}}}}}{{{2}\mu_{{{b}}}{g}}}}={\frac{{{\left({4.50}\frac{{m}}{{s}}\right)}^{{{2}}}}}{{{2}{\left({0.100}\right)}{\left({9.8}\frac{{m}}{{s}^{{{2}}}}\right)}}}}={10.3}{m}\)