 Suggested languages for you:

Europe

Answers without the blur. Sign up and see all textbooks for free! Q3.4-18E

Expert-verified Found in: Page 116 ### Fundamentals Of Differential Equations And Boundary Value Problems

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 is released from the top of an inclined plane that is inclined ${\mathbf{30}}{\mathbf{°}}$ to the horizontal (see Figure 3.11). Assume the gravitational force is constant, air resistance is negligible, and the coefficient of kinetic friction ${\mathbf{\mu }}{\mathbf{=}}{\mathbf{0}}{\mathbf{.}}{\mathbf{2}}$ . Determine the equation of motion for the object as it slides down the plane. If the top surface of the plane is 5 m long, what is the velocity of the object when it reaches the bottom?

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

See the step by step solution

## Step1: Find the value of velocity

There are two forces are written as,

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

Now put the given values then;

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

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

So, the values are written as;

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

## Step 2: Find the value of velocity by limits.

Now, find the value of velocity then,

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

## Step 3: Evaluate the equation of motion.

Now, for the value of x(t),

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

Therefore, the results are $x\left(t\right)=5.66 t+c$ and $v\left(t\right)=5.66 m/\mathrm{sec}$. ### Want to see more solutions like these? 