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Q3.4-18E

Expert-verifiedFound in: Page 116

Book edition
9th

Author(s)
R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages
616 pages

ISBN
9780321977069

**When an object slides on a surface, it encounters a resistance force called friction. This force has a magnitude of ${\mathbf{\mu N}}$** **, where ${\mathbf{\mu}}$** **the coefficient of kinetic friction and N is the magnitude of the normal force that the surface applies to the object. Suppose an object of mass ** 30 kg

The velocity of the object when it reaches the bottom is** $V\left(t\right)=5.66m/\mathrm{sec}$** and $x\left(t\right)=5.66t+c$ .

There are two forces are written as,

${F}_{1}=mg\u200a\mathrm{sin}\u200a{30}^{o}\phantom{\rule{0ex}{0ex}}{F}_{2}=-\mu \u200amg\u200a\mathrm{cos}\u200a{30}^{o}$

Now put the given values then;

$m\frac{dv}{dt}=mgsin{30}^{o}-\mu mgcos{30}^{o}\phantom{\rule{0ex}{0ex}}\frac{dv}{dt}=gsin{30}^{o}-\mu gcos{30}^{o}\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\left(\mathrm{Puttingthevaluesof}\u200a\mathrm{g}=0.2\right)\phantom{\rule{0ex}{0ex}}\frac{dv}{dt}=3.207\phantom{\rule{0ex}{0ex}}$

Since $v\left(t\right)=V\left(x\left(t\right)\right)$ ,

So, the values are written as;

$\frac{dv}{dt}=V\frac{dV}{dx}\phantom{\rule{0ex}{0ex}}V\frac{dV}{dx}=3.207\phantom{\rule{0ex}{0ex}}$

Now, find the value of velocity then,

$\int VdV=\underset{0}{\overset{5}{\int}}3.207dx\phantom{\rule{0ex}{0ex}}\frac{{V}^{2}}{2}-{\left.3.207x\right|}_{0}^{5}=0\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\u200a\left(\mathbf{Integratewithlimits}\mathbf{0}\mathbf{to}\mathbf{5}\right)\phantom{\rule{0ex}{0ex}}{V}^{2}=32.07\phantom{\rule{0ex}{0ex}}V\left(t\right)=5.66m/\mathrm{sec}\phantom{\rule{0ex}{0ex}}$

Now, for the value of x(t),

$v=5.66\phantom{\rule{0ex}{0ex}}\frac{dx}{dt}=5.66\phantom{\rule{0ex}{0ex}}x\u200a\left(t\right)=5.66\u200at+c\phantom{\rule{0ex}{0ex}}$

Therefore, the results are $x\left(t\right)=5.66\u200at+c$ and $v\left(t\right)=5.66\u200am/\mathrm{sec}$.

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