Q75P - University Physics with Modern Physics

Question

Question: You place a book of mass $\mathbf{5}.\mathbf{00}\mathbf{kg}$  against a vertical wall. You apply a constant force  $\overrightarrow{\mathbf{F}}$ to the book, where $\mathbf{F}\mathbf{=}\mathbf{96}\mathbf{.}\mathbf{0}\mathbf{N}$  and the force is at an angle of  $\mathbf{60}\mathbf{.}{\mathbf{0}}^{\mathbf{\circ }}$ above the horizontal (Fig. P5.75). The coefficient of kinetic friction between the book and the wall is 0.300 m . If the book is initially at rest, what is its speed after it has travelled  0.400 m up the wall?

Step-by-Step Solution

Answer

 Answer

The speed of the book is 1.74 m/s , after it had travelled  0,400m.

Step 1: Identification of the given data

The given data can be listed below as:

• The mass of the book is 500m .
• The force applied is N =96.0N .
• The angle subtended by the force with the horizontal is $\theta ={60}^0$ .
• The kinetic frictional coefficient is ${\mu }_K =0.3_{}$.

The distance travelled by the book is h = 4.00 m .

Step 2: Significance of the velocity

The velocity is described as the distance covered by a particular object in unit time. Velocity is also described as rate of change of displacement with respect to time.

Step 3: Determination of the speed

The vector diagram of the book has been drawn below:

 

The frictional force   is acting in the downwards direction along with the weight of the body  . The normal reaction force   is acting in the left direction. The x and y-component of the force E   are    $F\cos 60° and F\sin 6 0°$  acting along -x and +y-axis respectively.

 

From the above diagram, the net force acting in along x-axis is given as:

 

  $n=F\cos 60° ······················\left(1\right)$

 

Here,   is the normal reaction force and   is the magnitude of force acting on the book.

 

From the above diagram, the net force acting in along y-axis is given as:

 $F\sin 60°-mg-f_k=ma ····························\left(2\right)$

 

 

Here, m  is the mass of the book,  g is the acceleration due to gravity,  $f_k$ is the frictional force and  a is the net acceleration of the book.

 

The equation of the frictional force is expressed as:

 $f_k={\mu }_{k}n$

  

 

Here${\mu }_k$  is the static frictional coefficient.

 

Substituting the above value in the equation (2), we get

data-custom-editor="chemistry" $F\sin 60°-mg-{\mu }_{k}n=ma ····························\left(3\right)$

 

 Comparing equation (1) and equation (3), we get the following relation-

 $$\begin{gathered} F\sin 60°-mg-{\mu }_{k}F\cos 60°=ma \\ a=\frac{F\sin 60°-mg-{\mu }_{k}F\cos 60°}{m} \end{gathered}$$

 

 

Substituting given values for the variables in above equation.

 $$\begin{gathered} a=\frac{\left(96.0 \text{N}\right)\left(0.86\right)-\left(5.00 \text{kg}\right)\left(9.8 {\text{m/s}}^{\text{2}}\right)-\left(0.3\right)\left(96.0 \text{N}\right)\left(0.5\right)}{\left(5.00 \text{kg}\right)} \\ =\frac{\left(82.56 \text{N}\times \frac{1 \text{kg}·{\text{m/s}}^{\text{2}}}{1 \text{N}}\right)-\left(49 \text{kg}·{\text{m/s}}^{\text{2}}\right)-\left(14.4 \text{N}\times \frac{1 \text{kg}·{\text{m/s}}^{\text{2}}}{1 \text{N}}\right)}{\left(5.00 \text{kg}\right)} \\ =\frac{\left(82.56 \text{kg}·{\text{m/s}}^{\text{2}}\right)-\left(49 \text{kg}·{\text{m/s}}^{\text{2}}\right)-\left(14.4 \text{kg}·{\text{m/s}}^{\text{2}}\right)}{\left(5.00 \text{kg}\right)} \\ =3.83 {\text{m/s}}^{\text{2}} \end{gathered}$$

 

 

Using the third equation of motion-

 $$\begin{gathered} v^2=u^2+2ah \\ v=\sqrt{u^2+2ah} \end{gathered}$$

  

 

Here, v is the final speed and u  is the initial speed of the book.   is the distance travelled by the book.

 

As the book was at rest initially, then the initial speed of the book is zero.

 

Substitute the values in the above equation.

 $$\begin{gathered} v=\sqrt{0+2\left(3.8 {\text{m/s}}^{\text{2}}\right)\left(0.4 \text{m}\right)} \\ =\sqrt{\left(7.6 {\text{m/s}}^{\text{2}}\right)\left(0.4 \text{m}\right)} \\ =\sqrt{\left(3.04 {\text{m}}^2{\text{/s}}^{\text{2}}\right)} \\ =1.74 \text{m/s} \end{gathered}$$

 

 

Thus, the speed of the book is 1.74 m/s .