Chapter 5, Problem 71A

(a)

To determine

The acceleration of the block when the surface is frictionless.

Answer to Problem 71A

The acceleration of the block when the surface is frictionless is $8{\text{m/s}}^2$ .

Explanation of Solution

Given:

The mass $\left(m\right)$ of the block is $5.0\text{kg}$ .

The applied force $\left(F_{\text{applied}}\right)$ exerted on the block is $40.0\text{N}$ .

The net acceleration $\left(a_{\text{net}}\right)$ on the block is $6.0{\text{m/s}}^2$ .

Consider the acceleration due to gravity $\left(g\right)$ of Earth is $9.80{\text{m/s}}^2$ .

Formula used:

Newton’s Second Law:

The object’s acceleration is equivalent to the sum of the forces acting on the object divided by the object’s mass. That is,

  $\begin{array}{l}a=\frac{F_{\text{net}}}{m}\\ F_{\text{net}}=ma\end{array}$

Calculation:

The surface is frictionless. So, the frictional force acting on the block is denoted by $f=0\text{N}$ .

The acceleration of the block is denoted by $a$ .

Apply Newton’s second law of motion,

  $\begin{array}{l}{F}_{\text{net}}=F_{\text{applied}}-f\\ ma=F_{\text{applied}}-0\\ a=\left(\frac{F_{\text{applied}}}{m}\right)\end{array}$

Substitute the values for the applied force and the mass of the block.

  $\begin{array}{l}a=\left\{\frac{40\text{N}\times \left(\frac{\text{1}\text{kg}\cdot {\text{m/s}}^2}{1\text{N}}\right)}{5\text{kg}}\right\}\\ a=\left(\frac{40\text{kg}\cdot {\text{m/s}}^2}{5\text{kg}}\right)\\ a=8{\text{m/s}}^2\end{array}$

Conclusion:

Therefore, the acceleration of the block when the surface is frictionless is $8{\text{m/s}}^2$ .

(b)

To determine

The kinetic friction force acting on the block.

Answer to Problem 71A

The kinetic friction force acting on the block is $10\text{N}$ .

Explanation of Solution

Given:

The mass $\left(m\right)$ of the block is $5.0\text{kg}$ .

The applied force $\left(F_{\text{applied}}\right)$ exerted on the block is $40.0\text{N}$ .

The net acceleration $\left(a_{\text{net}}\right)$ on the block is $6.0{\text{m/s}}^2$ .

Consider the acceleration due to gravity $\left(g\right)$ of Earth is $9.80{\text{m/s}}^2$ .

Formula used:

Consider a horizontal force acts on the object and the object slides over the surface, frictional force acts between the object and surface. The expression for the kinetic frictional force between the object and the surface as follows:

  $f_{\text{k}}={\mu }_{\text{k}}N$

Here, $N$ is normal force acting on the object and ${\mu }_{\text{k}}$ is the coefficient of kinetic friction.

Newton’s Second Law:

The object’s acceleration is equivalent to the sum of the forces acting on the object divided by the object’s mass. That is,

  $\begin{array}{l}a=\frac{F_{\text{net}}}{m}\\ F_{\text{net}}=ma\end{array}$

Calculation:

Consider the kinetic friction force acting on the block is denoted by $f$ .

The relation between the applied force on the block and the frictional resistance is:

  $\begin{array}{l}{F}_{\text{net}}=F_{\text{applied}}-f\\ m{a}_{\text{net}}=F_{\text{applied}}-f\\ f=F_{\text{applied}}-m{a}_{\text{net}}\end{array}$

Substitute the values for the applied force, mass of the block and net acceleration.

  $\begin{array}{l}{F}_{\text{net}}=F_{\text{applied}}-f\\ m{a}_{\text{net}}=F_{\text{applied}}-f\\ f=F_{\text{applied}}-m{a}_{\text{net}}\\ f=40\text{N}-5\text{kg}\times \text{6}\text{.0}{\text{m/s}}^2\\ f=40\text{N}-30\text{kg}\cdot {\text{m/s}}^2\times \left(\frac{1\text{N}}{\text{1}\text{kg}\cdot {\text{m/s}}^2}\right)\\ f=40\text{N}-30\text{N}\\ f=10\text{N}\end{array}$

Conclusion:

Therefore, the kinetic friction acting on the block is $10\text{N}$ .

(c)

To determine

The coefficient of kinetic friction between the block and the surface.

Answer to Problem 71A

The coefficient of kinetic friction between the block and the surface is ${\mu }_{\text{k}}=0.204$ .

Explanation of Solution

Given:

The mass $\left(m\right)$ of the block is $5.0\text{kg}$ .

The applied force $\left(F_{\text{applied}}\right)$ exerted on the block is $40.0\text{N}$ .

The net acceleration $\left(a_{\text{net}}\right)$ on the block is $6.0{\text{m/s}}^2$ .

Consider the acceleration due to gravity $\left(g\right)$ of Earth is $9.80{\text{m/s}}^2$ .

Formula used:

Consider a horizontal force acts on the object and the object slides over the surface, frictional force acts between the object and surface. The expression for the kinetic frictional force between the object and the surface as follows:

  $f_{\text{k}}={\mu }_{\text{k}}N$

Here, $N$ is normal force acting on the object and ${\mu }_{\text{k}}$ is the coefficient of kinetic friction.

Calculation:

The weight $\left(F_{\text{g}}\right)$ of the block is,

  $\begin{array}{c}{F}_{\text{g}}=mg\\ =5\text{kg}\times \text{9}\text{.80}{\text{m/s}}^2\\ =49\text{kg}\cdot {\text{m/s}}^2\times \left(\frac{1\text{N}}{1\text{kg}\cdot {\text{m/s}}^2}\right)\\ =49\text{N}\end{array}$

The normal force $\left(N\right)$ acting on the block is equal to the weight of the block.

  $\begin{array}{c}N=F_{\text{g}}\\ N=49\text{N}\end{array}$

From part (b), the friction force $\left(f_{\text{k}}\right)$ acting on the block is $10\text{N}$ .

So, the value of the coefficient of kinetic friction $\left({\mu }_{\text{k}}\right)$ as follows:

  $\begin{array}{l}{f}_{\text{k}}={\mu }_{\text{k}}N\\ 10\text{N}={\mu }_{\text{k}}\left(49\text{N}\right)\\ {\mu }_{\text{k}}=\left(\frac{10\text{N}}{49\text{N}}\right)\\ {\mu }_{\text{k}}=0.204\end{array}$

Conclusion:

Therefore, the coefficient of kinetic friction between the block and the surface is ${\mu }_{\text{k}}=0.204$ .