# A box is sliding with a speed of 4.50 m/s on a horizontal surface when, at point P, it encounters a rough section. On the rough section, the coefficient of friction is not constant but starts at .100 at P and increases linerly with distance past P, reaching a value of .600 at 12.5 m past point P. (a) Use the work energy theorem to find how far this box slides before stopping. (b) What is the coefficient of friction at the stopping point? (c) How far would the box have slid iff the friciton coefficient didn't increase, but instead had the constant value of .1?

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A box is sliding with a speed of 4.50 m/s on a horizontal surface when, at point P, it encounters a rough section. On the rough section, the coefficient of friction is not constant but starts at .100 at P and increases linerly with distance past P, reaching a value of .600 at 12.5 m past point P. (a) Use the work energy theorem to find how far this box slides before stopping. (b) What is the coefficient of friction at the stopping point? (c) How far would the box have slid iff the friciton coefficient didn't increase, but instead had the constant value of .1?

2021-03-24
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)
$$\displaystyle{W}_{{{f}}}={K}{E}_{{{f}}}-{K}{E}_{{{i}}}$$
$$\displaystyle-\mu_{{{b}}}{m}{g}{x}={0}-{\frac{{{1}}}{{{2}}}}{m}{{v}_{{{1}}}^{{{2}}}}$$
$$\displaystyle{x}={\frac{{{{v}_{{{1}}}^{{{2}}}}}}{{{2}\mu_{{{b}}}{g}}}}={\frac{{{\left({4.50}\frac{{m}}{{s}}\right)}^{{{2}}}}}{{{2}{\left({0.100}\right)}{\left({9.8}\frac{{m}}{{s}^{{{2}}}}\right)}}}}={10.3}{m}$$

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