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Found in: Page 140

### Fundamentals Of Physics

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${\mathbit{\theta }}{\mathbf{}}{\mathbf{=}}{\mathbf{20}}{\mathbf{°}}$ .What is the magnitude of the acceleration of the block across the floor if the friction coefficients are (a) ${\mathbit{\mu }}{\mathbit{s}}{\mathbf{}}{\mathbf{=}}{\mathbf{0}}{\mathbf{.}}{\mathbf{600}}$and${{\mathbit{\mu }}}_{{\mathbf{k}}}{\mathbf{=}}{\mathbf{0}}{\mathbf{.}}{\mathbf{500}}$ and (b)${\mathbit{\mu }}{\mathbit{s}}{\mathbf{}}{\mathbf{=}}{\mathbf{0}}{\mathbf{.}}{\mathbf{400}}$ and${{\mathbit{\mu }}}_{{\mathbf{k}}}{\mathbf{=}}{\mathbf{0}}{\mathbf{.}}{\mathbf{300}}$ ?

(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.$\frac{\text{m}}{{\text{s}}^{\text{2}}}$

See the step by step solution

## Step 1: Given

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

## Step 2: Determining the concept and the formula:

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.

## Step 3: (a) Determinethe magnitude of the acceleration of the block across the floor if the friction coefficients are μs =0.600 and μk =0.500

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

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

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 .$\frac{\text{m}}{{\text{s}}^{\text{2}}}$

## Step 4: (b) Determining the magnitude of the acceleration of the block across the floor if the friction coefficients are μs =0.400andμk=0.300

Static frictional force on the block from the floor is,

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

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,

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