A stockroom worker pushes a box with mass 16.8 kg on a horizontal surface with a constant speed of 3.50 m/s. The coefficient of kinetic friction between the box and the surface is 0.20. (a) What horizontal force must the worker apply to maintain the motion? (b) If the force calculated in part (a) is removed, how far does the box slide before coming to rest?

The given data is listed below as,

The coefficient of friction is the amount of friction that occurs when two surfaces are in contact with each other. If the coefficient of friction has a low value, one surface can easily slide over the other.

(a)

Using the Newton’s second law of motion,

$F-F_f=ma$

When a = 0

$\begin{array}{l}F=F_f\\ =\mu N\\ =\mu mg\end{array}$

Here $\mu$is the coefficient of friction, m is the mass of the box, and g is the acceleration due to gravity.

Substitute 0.20 for $\mu$, 16.8 kg for m,$9.8m/s^2$ for g in the above equation.

$$\begin{gathered} F_f=\left(0.20\right)\left(16.8kg\right)\left(9.8m/s^2\right) \\ =32.928N \end{gathered}$$

Hence, the required horizontal force is 32.928 N.

(b)

Using the Newton’s second law of motion,

$m=F-F_f$

When, F = 0

$\begin{array}{ll}ma=-F_f& \\ ma=-\mu mg& \\ a=-\mu g& \end{array}$

Substitute 0.20 for $\mu$,$9.8m/s^2$ for g in the above equation.

$$\begin{gathered} a=-(0.20)\hspace{0.33em}(9.8m/s^2) \\ a=-1.96\hspace{0.33em}m/s^2 \end{gathered}$$

The equation of motion is expressed as,

$s=\frac{v^2-v_o^2}{2a}$

Here v is the final velocity,$v_{\circ }$ is the initial velocity, a is the acceleration.

Substitute 0 for v, 3.50 m/s for $v_{\circ }$, $-1.96m/s^2$for in the above equation.

$\begin{array}{l}s=\frac{(0{)}^2-{(3.50m/s)}^2}{2(-1.9\mathrm{m}/{\mathrm{s}}^2)}\\ =3.1m\end{array}$

Hence, the required distance is 3.1 m.