A person pushes horizontally with a force of$\mathbf{220}\mathbf{}\mathit{N}$on a$\mathbf{55}\mathbf{}\mathit{k}\mathit{g}$crate to move it across a level floor. The coefficient of kinetic friction between the crate and the floor is$\mathbf{0}\mathbf{.}\mathbf{35}$. What is the magnitude of (a) the frictional force and (b) the acceleration of the crate?

Mass of the crate,$M=55Kg$

Horizontal force,$F=220\text{N}$

Coefficient of kinetic friction force,${\mu }_k=0.35$

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. With this the frictional force and the acceleration can be calculated.

Consider the formula for the net force as:

$F_{net}=\sum ma$

Here, F is the net force, mis mass and ais an acceleration.

Free body diagram of the crate:

By using Newton’s 2nd law along y direction,

$\sum _{}^{}{F}_y=m{a}_y$

Since crate is not moving along y, $a_y=0$.

$$\begin{gathered} N-mg=0 \\ N=mg \end{gathered}$$

Relation between frictional force and normal force is

Hence, the magnitude of the frictional force is 188.84 N.

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

$\sum _{}^{}{F}_x=m{a}_x$

Since crate is moving along x, let$a_x=a$

$F-f_s=ma$

Rewrite the formulas and substitute the values and solve as:

$\begin{array}{rcl}a& =& \frac{F-f_s}{m}\\ & =& \frac{220-188.84}{55}\\ & =& 0.5665\frac{\text{m}}{{\text{s}}^{\text{2}}}\\ & & \end{array}$

Hence. the magnitude of the acceleration of the crate is $0.5665\frac{\text{m}}{{\text{s}}^{\text{2}}}$.