Q27E - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

1. The horizontal force is 32.928N. 
2. The distance covered by box is 3.1 m.

Step 1: Identification of the given data

The given data is listed below as,

 

• Mass of the box is m = 16.8 kg.

 

• Velocity of the box is $v_{\circ }=3.50 m/s$.

 

• The coefficient of friction is $\mu =0.20$.

Step 2: Significance of coefficient of friction

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.

Step 3: Determination of the horizontal force must the worker apply to maintain the motion

(a)

 

Using the Newton’s second law of motion,

 

$F-F_f=ma$

 

When a = 0

 

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

 

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, $9.8 m/s^2$ for g in the above equation.

 

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

 

Hence, the required horizontal force is 32.928 N.

Step 4: Determination of distance of box slide before it coming to rest

(b)

 

Using the Newton’s second law of motion,

 

$m=F-F_f$

 

When,  F = 0

 

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

 

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

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

 

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

 

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

 

Hence, the required distance is 3.1 m.