Q66P - University Physics with Modern Physics

Question

A box is sliding with a constant speed of 4.00m/s in the +x -direction on a horizontal, frictionless surface. At x = 0 the box encounters a rough patch of the surface, and then the surface becomes even rougher. Between x = 0 and x = 2.00 m, the coefficient of kinetic friction between the box and the surface is 0.200; between x = 2.00 m and x = 4.00 m , it is 0.400 . (a) What is the  -coordinate of the point where the box comes to rest? (b) How much time does it take the box to come to rest after it first encounters the rough patch at x = 0 ?

Step-by-Step Solution

Answer

 (a) The x  coordinate of the point is 3.04 m .

(b) The time taken by the box to come to rest is 1.31 s.

Step 1: Identification of the given data:

The given data can be listed below as:

• The velocity of the box is u = 4.00 m/s .
• The kinetic friction’s coefficient between the surface and the box in the first case is ${\mathrm{\mu }}_1=0.200$ .
• The kinetic friction’s coefficient between the surface and the box in the second case is ${\mathrm{\mu }}_2=0.400$ .

Step 2: Significance of the friction:

Friction is described as the resistance that one object or surface encounters on another object. The coefficient of friction is described as the division between the friction and the normal force.

Step 3: (a) Determination of the x coordinate of the point:

The free-body diagram of the box has been drawn below:

From the above diagram, the equation of the summation of the force in the   direction is expressed as:

 

${\mathrm{ma}}_{\mathrm{x}}+{\mathrm{f}}_{\mathrm{k}}=0$                                                                                                                ….. (1)

 

Here, m is the mass of the box, ${\mathrm{a}}_{\mathrm{x}}$ is the acceleration in the x direction and ${\mathrm{f}}_{\mathrm{k}}$ is the frictional force.

 

From the above diagram, the equation of the summation of the force in the   direction is expressed as:

 

n = mg                                                                                                                      ….. (2)

 

Here, n is the normal force and g is the acceleration due to gravity.

 

The equation of the frictional force is expressed as:

 

${\mathrm{f}}_{\mathrm{k}}=\mathrm{\mu n}$ 

 

Here, $\mu$is the coefficient of the frictional force.

 

Substitute mg for n in the above equation.

 

${\mathrm{f}}_{\mathrm{k}}=\mathrm{\mu mg}$

Substitute the value of the above equation in the equation (i).

 

$$\begin{gathered} -\mathrm{\mu mg}={\mathrm{ma}}_{\mathrm{x}} \\ {\mathrm{a}}_{\mathrm{x}}=-\mathrm{\mu g} \end{gathered}$$ 

                                                                                                                    ….. (3)

In the first case,

 

Substitute 0.200 for $\mathrm{\mu }$ and $9.8 \mathrm{m}/{\mathrm{s}}^2$ for g into equation (3).

$$\begin{gathered} {\mathrm{a}}_1=\left(0.200\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right) \\ =-1.96\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$   

In the second case,

 

Substitute 0.400  for $\mu$ and $9.8 \mathrm{m}/{\mathrm{s}}^2$ for into equation (3).

 

$$\begin{gathered} {\mathrm{a}}_2=\left(0.400\right)\left(9.8 \mathrm{m}/{\mathrm{s}}^2\right) \\ =-3.92 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$ 

 

The equation of the final velocity is expressed as:

 

${\mathrm{v}}^2={\mathrm{u}}^2+2{\mathrm{a}}_1\mathrm{x}$

Here,  v is the final and u is the initial velocity of the box. ${\mathrm{a}}_1$  is the acceleration of the box in the first case and   is the initial distance moved by the box.

 

Substitute  4.00 m/s for u , - 1.96 $\mathrm{m}/{\mathrm{s}}^2$ for ${\mathrm{a}}_1$ and  2.00 m for x  in the above equation.

 

${\mathrm{v}}^2={\mathrm{u}}^2+2{\mathrm{a}}_1\mathrm{x}$

Here,   is the final and   is the initial velocity of the box.   is the acceleration of the box in the first case and   is the initial distance moved by the box.

 

Substitute 4.00 m/s for $\mathrm{u}-1.96\mathrm{m}/{\mathrm{s}}^2$, for $a_1$ and 2.00 m for x  in the above equation.

 

$$\begin{gathered} {\mathrm{v}}^2={\left(4.00\mathrm{m}/\mathrm{s}\right)}^2+2\left(-1.96 \mathrm{m}/{\mathrm{s}}^2\right)\left(2.00 \mathrm{m}\right) \\ =16.00{\mathrm{m}}^2/{\mathrm{s}}^2-\left(7.84 \mathrm{m}/{\mathrm{s}}^2\right) \\ =8.16{\mathrm{m}}^2/{\mathrm{s}}^2 \\ \mathrm{v}=2.86\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

The equation of the x coordinate of the point is expressed as:

 

$$\begin{gathered} {\mathrm{v}}^2={\mathrm{u}}^2+2{\mathrm{a}}_2\left(\mathrm{x}-{\mathrm{x}}_1\right) \\ {\mathrm{v}}^2={\mathrm{u}}^{2}2{\mathrm{a}}_2\left(\mathrm{x}-{\mathrm{x}}_1\right) \\ \left(\mathrm{x}-{\mathrm{x}}_1\right)=\frac{{\mathrm{v}}^2-{\mathrm{u}}^2}{2{\mathrm{a}}_2} \\ \mathrm{x}={\mathrm{x}}_1-\frac{{\mathrm{v}}^2-{\mathrm{u}}^2}{2{\mathrm{a}}_2} \end{gathered}$$  

 

Here, x is described as the x coordinate of the point, $x_1$ is the initial distance moved by the box and $x_2$ is the acceleration of the box in the second case.

 

Substitute the values in the above equation.

$$\begin{gathered} \mathrm{x}=2.00\mathrm{m}-\frac{{\left(2.86\mathrm{m}/\mathrm{s}\right)}^2}{2\left(-3.92\mathrm{m}/{\mathrm{s}}^2\right)} \\ =2.00\mathrm{m}+\frac{\left(8.17{\mathrm{m}}^2/{\mathrm{s}}^2\right)}{\left(7.84\mathrm{m}/{\mathrm{s}}^2\right)} \\ =2.00\mathrm{m}+1.04 \mathrm{m} \\ =3.04 \mathrm{m} \end{gathered}$$

Thus, the x coordinate of the point is 3.04 m .

Step 4: (b) Determination of the time:

The equation of the final time is expressed as:

 

$$\begin{gathered} {\mathrm{v}}_1=\mathrm{v}+{\mathrm{a}}_2{\mathrm{t}}_2 \\ {\mathrm{t}}_2=\frac{\mathrm{v}-{\mathrm{v}}_1}{{\mathrm{a}}_2} \end{gathered}$$  

 

Here, $t_2$ is the final time.

 

Substitute the values in the above equation.

 

$$\begin{gathered} {\mathrm{t}}_2=\frac{0-2.86 \mathrm{m}/{\mathrm{s}}^2}{-1.96 \mathrm{m}/{\mathrm{s}}^2} \\ =\frac{-2.86 \mathrm{m}/{\mathrm{s}}^2}{-1.96 \mathrm{m}/{\mathrm{s}}^2} \\ =0.730 \mathrm{s} \end{gathered}$$ 

 

The equation of the initial time is expressed as:

 

 $$\begin{gathered} \mathrm{v}=\mathrm{u}+{\mathrm{a}}_1{\mathrm{t}}_1 \\ {\mathrm{t}}_1=\frac{\mathrm{v}-\mathrm{u}}{{\mathrm{a}}_1} \end{gathered}$$

 

Here, ${\mathrm{t}}_1$ is the initial time.

 

Substitute the values in the above equation.

 

$$\begin{gathered} {\mathrm{t}}_1=\frac{2.86 \mathrm{m}/{\mathrm{s}}^2-4.00 \mathrm{m}/{\mathrm{s}}^2}{-1.96 \mathrm{m}/{\mathrm{s}}^2} \\ =\frac{-1.14 \mathrm{m}/{\mathrm{s}}^2}{-1.96 \mathrm{m}/{\mathrm{s}}^2} \\ =0.582 \mathrm{s} \end{gathered}$$ 

 

The equation of the total time is expressed as:

 

$\mathrm{t}={\mathrm{t}}_1+{\mathrm{t}}_2$ 

 

Substitute the values in the above equation.

 

 $$\begin{gathered} \mathrm{t}=0.582 \mathrm{s}+0.730 \mathrm{s} \\ =1.31 \mathrm{s} \end{gathered}$$

 

Thus, the time taken by the box to come to rest is 1.31 s .