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Q10P

Expert-verifiedFound in: Page 140

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**Figure 6-20 shows an initially stationary block of mass ****on a floor. A force of magnitude ****is then applied at upward angle${\mathit{\theta}}{\mathbf{}}{\mathbf{=}}{\mathbf{20}}{\mathbf{\xb0}}$ ****.What is the magnitude of the acceleration of the block across the floor if the friction coefficients are (a) **** ${\mathit{\mu}}{\mathit{s}}{\mathbf{}}{\mathbf{=}}{\mathbf{0}}{\mathbf{.}}{\mathbf{600}}$and${{\mathit{\mu}}}_{{\mathbf{k}}}{\mathbf{=}}{\mathbf{0}}{\mathbf{.}}{\mathbf{500}}$ ****and (b)${\mathit{\mu}}{\mathit{s}}{\mathbf{}}{\mathbf{=}}{\mathbf{0}}{\mathbf{.}}{\mathbf{400}}$**** and${{\mathit{\mu}}}_{{\mathbf{k}}}{\mathbf{=}}{\mathbf{0}}{\mathbf{.}}{\mathbf{300}}$ ****?**

**(a)** The magnitude of the acceleration of the block across the floor is 0 .$\frac{\text{m}}{{\text{s}}^{\text{2}}}$

**(b)** The magnitude of the acceleration of the block across the floor is 2.15.$\frac{\text{m}}{{\text{s}}^{\text{2}}}$

Mass of the block, m

Force on the block, $F=0.500mg.N$

Angle, $\theta ={20}^{0}$

${\mu}_{s}=$0.600 and ${\mu}_{K}=$ 0.500

${\mu}_{s}=$0.400 and ${\mu}_{K}=$ 0.300

**The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. Use the Newton's 2nd law of motionalong vertical and horizontal direction. **

Write the formula for the net force:

${F}_{net}=\sum ma$

Here, F is the net force, m is mass and a is an acceleration.

** **

By using Newton’s 2nd law along y direction,block is not moving along y. So, the acceleration is 0.

$\sum _{}^{}{F}_{y}=0$

$N+F\mathrm{sin}\left(20\right)-mg=0$

Substitute the values and solve as:

$N=mg-\left(0.500\text{mg}\right)\mathrm{sin}\left(20\right)\phantom{\rule{0ex}{0ex}}=0.83mgN$

Static frictional force on the block from the floor is,

$\begin{array}{rcl}{f}_{s}& =& {\mu}_{s}N\\ & =& \left(0.600\right)\left(0.83\text{mg}\right)\\ & =& 0.50\text{mgN}\\ & & \end{array}$

Opposite force to the ${f}_{s}$ is,

$\begin{array}{rcl}F\mathrm{cos}\left(20\right)& =& \left(0.500\text{mg}\right)\mathrm{cos}\left(20\right)\\ & =& 0.47\text{mg}\xb7\text{N}\\ & & \end{array}$

So, here, $F\mathrm{cos}\left(20\right)<{f}_{s}$therefore block cannot move.

Therefore, the magnitude of the acceleration of the block across the floor is 0 .$\frac{\text{m}}{{\text{s}}^{\text{2}}}$

Static frictional force on the block from the floor is,

$\begin{array}{rcl}{f}_{s}& =& {\mu}_{s}N\\ & =& \left(0.400\right)\left(0.83\text{mg}\right)\\ & =& 0.33\text{mg}\xb7\text{N}\\ & & \end{array}$

Opposite force to the ${f}_{s}$ is,

$\begin{array}{rcl}F\mathrm{cos}\left(20\right)& =& \left(0.500mg\right)\mathrm{cos}\left(20\right)\\ & =& 0.47mg.N\end{array}$

So,here . $F\mathrm{cos}\left(20\right)>{f}_{s}$Therefore, the block will be moved.

Kinetic frictional force on the block is,

$\begin{array}{rcl}{f}_{k}& =& {\mu}_{k}N\\ & =& \left(0.300\right)\left(0.83\text{mg}\right)\\ & =& 0.25mg.N\end{array}$

By using Newton’s 2nd law along x direction is,

$\sum _{}^{}{F}_{x}=m{a}_{x}\phantom{\rule{0ex}{0ex}}F\mathrm{cos}\left(20\right)-{f}_{k}=m{a}_{x}\phantom{\rule{0ex}{0ex}}\left(0.500mg\right)\mathrm{cos}\left(20\right)-\left(0.25mg\right)=m{a}_{x}\phantom{\rule{0ex}{0ex}}{a}_{x}=2.15\frac{\text{m}}{{\text{s}}^{\text{2}}}$

Therefore, the magnitude of the acceleration of the block across the floor is 2.15.$\frac{\text{m}}{{\text{s}}^{\text{2}}}$

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