Q54P - University Physics with Modern Physics

Question

An 8.00 kg box sits on a level floor. You give the box a sharp push and find that it travels 8.22 m in 2.8 s
 before coming to rest again. (a) You measure that with a different push the box traveled 4.20 m in 2.0 s. Do you think the box has a constant acceleration as it slows down? Explain your reasoning. (b) You add books to the box to increase its mass. Repeating the experiment, you give the box a push and measure how long it takes the box to come to rest and how far the box travels. The results, including the initial experiment with no added mass, are given in the table:

               

In each case, did your push give the box the same initial speed? What is the ratio between the greatest initial speed and the smallest initial speed for these four cases? (c) Is the average horizontal force f exerted on the box by the floor the same in each case? Graph the magnitude of force f versus the total mass m of the box plus its contents, and use your graph to determine an equation for f as a function of m.

Step-by-Step Solution

Answer

(a) The box has constant acceleration as it slows down.

(b) The box has not been pushed with the same initial speed in each case.

The ratio between the greatest and the smallest initial speed is $1.2$.

(c) The average horizontal force exerted by the floor on the box is not same.

                   

The equation is $\mathrm{F}=\mathrm{m}\left(2.1\mathrm{m}/{\mathrm{s}}^2\right)$ .  

Step 1: Identification of the given data:

The given data can be listed below as:

• The mass of the box is, $\mathrm{m}=8.00 \mathrm{kg}$ .
• The distance travelled by the box initially is $\mathrm{s}=8.00 \mathrm{m}$.
• The time taken by the box to travel to that distance initially is $\mathrm{t}=2.8 \mathrm{s}$.
• The distance travelled by the box finally is ${\mathrm{s}}_1=4.20 \mathrm{m}$.
• The time taken by the box to travel to that distance finally is ${\mathrm{t}}_1=2.0 \mathrm{s}$.

Step 2: Significance of Newton’s second law:

Newton’s second law states that the force applied to the object is directly proportional to the mass and the acceleration of that object. The second law also provides an idea of the net force exerted by the object.

Step 3: (a) Determination of the acceleration of the box:

The equation of the initial speed of the box is expressed as:

 

 ${\mathrm{v}}_{\mathrm{f}}={\mathrm{v}}_{\mathrm{i}}+\mathrm{at}$ 

 

Here, ${\mathrm{v}}_{\mathrm{f}}$ is the final and ${\mathrm{v}}_{\mathrm{i}}$ is the initial speed, a is the acceleration of the box and $\mathrm{t}$ is the time taken by the box to travel to that distance initially.

 

As finally the box came to rest, then the final speed of the box is zero.

 

Substitute 0 for $v_f$ in the above equation.

 

 $$\begin{gathered} 0={\mathrm{v}}_{\mathrm{i}}+\mathrm{at} \\ {\mathrm{v}}_{\mathrm{i}}=-\mathrm{at} \end{gathered}$$

                                                                                                                     ….. (1)

 

The equation of the distance travelled by the box is expressed as:

 

 $\mathrm{x}={\mathrm{v}}_{\mathrm{i}}\mathrm{t}+\frac{1}{2}{\mathrm{at}}^2$

 

Here, $\mathrm{x}$ is the distance the box has travelled in the second case.

 

Substitute $-\mathrm{at}$ for ${\mathrm{v}}_{\mathrm{i}}$ in the above equation.

 

$$\begin{gathered} \mathrm{x}={\mathrm{at}}^2+\frac{1}{2}{\mathrm{at}}^2 \\ \mathrm{x}=-\frac{1}{2}{\mathrm{at}}^2 \end{gathered}$$ 

                                                                                                                 ….. (2)

 

The above equation can also be written as:

 $\mathrm{a}=-\frac{2\mathrm{x}}{{\mathrm{t}}^2}$

                                                                                                                 ….. (3)

 

Substitute $8.22 \mathrm{m}$ for $\mathrm{x}$ and $2.8 \mathrm{s}$ for $\mathrm{t}$ in the above equation.

 

$$\begin{gathered} \mathrm{a}=-\frac{2\left(8.22 \mathrm{m}\right)}{{\left(2.8 \mathrm{s}\right)}^2} \\ =-\frac{\left(16.44 \mathrm{m}\right)}{\left(7.84 {\mathrm{s}}^2\right)} \\ =-2.1 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$ 

 

Substitute $-2.1 \mathrm{m}/{\mathrm{s}}^2$ for $\mathrm{a}$ and $2 \mathrm{s}$ for $\mathrm{t}$ in the above equation.

 

 $$\begin{gathered} \mathrm{x}=-\frac{1}{2}\left(-2.1 \mathrm{m}/{\mathrm{s}}^2\right){\left(2 \mathrm{s}\right)}^2 \\ =-\frac{1}{2}\left(-2.1 \mathrm{m}/{\mathrm{s}}^2\right){\left(4 \mathrm{s}\right)}^2 \\ =\left(1.05 \mathrm{m}/{\mathrm{s}}^2\right){\left(4 \mathrm{s}\right)}^2 \\ =4.2 \mathrm{m} \end{gathered}$$

 

As the given travelled distance from the table matches the distance gathered, then it can be stated that the box has constant acceleration.

Thus, the box has constant acceleration as it slows down.

Step 4: (b) Determination of the initial speed of the box:

The equation (1) has been recalled below:

 ${\mathrm{v}}_{\mathrm{i}}=-\mathrm{at}$ 

Substitute $-\frac{2\mathrm{x}}{{\mathrm{t}}^2}$ for  a in the above equation.

 $$\begin{gathered} {\mathrm{v}}_{\mathrm{i}}=\frac{2\mathrm{x}}{{\mathrm{t}}^2}\mathrm{t} \\ =\frac{2\mathrm{x}}{\mathrm{t}} \end{gathered}$$

 

According to the table, as there are different distances along with different times taken by the box to move to a certain distance, hence the initial speeds are different.

 

For the first case,

 

Substitute $8.22 \mathrm{m}$ for $\mathrm{x}$ and $2.8 \mathrm{s}$ for $\mathrm{t}$ in the above equation.

 $$\begin{gathered} {\mathrm{v}}_{\mathrm{i}}=\frac{2\left(8.22 \mathrm{m}\right)}{\left(2.8 \mathrm{s}\right)} \\ =\frac{16.44 \mathrm{m}}{2.8 \mathrm{s}} \\ =5.87 \mathrm{m}/\mathrm{s} \end{gathered}$$

 

For the second case,

 

Substitute $10.75 \mathrm{m}$ for $\mathrm{x}$ and $3.2 \mathrm{s}$ for $\mathrm{t}$ in the above equation.

 

$$\begin{gathered} {\mathrm{v}}_{\mathrm{i}}=\frac{2\left(10.75 \mathrm{m}\right)}{\left(3.2 \mathrm{s}\right)} \\ =\frac{21.5 \mathrm{m}}{3.2 \mathrm{s}} \\ =6.72 \mathrm{m}/\mathrm{s} \end{gathered}$$ 

 

For the third case,

 

Substitute $9.45 \mathrm{m}$ for $\mathrm{x}$ and $3.2 \mathrm{s}$ for $\mathrm{t}$ in the above equation.

 $$\begin{gathered} {\mathrm{v}}_{\mathrm{i}}=\frac{2\left(9.45 \mathrm{m}\right)}{\left(3 \mathrm{s}\right)} \\ =\frac{18.9 \mathrm{m}}{3.2 \mathrm{s}} \\ =6.3 \mathrm{m}/\mathrm{s} \end{gathered}$$

 

For the fourth case,

 

Substitute $7.10 \mathrm{m}$ for $\mathrm{x}$ and $2.6 \mathrm{s}$ for $\mathrm{t}$ in the above equation.

 

 $$\begin{gathered} {\mathrm{v}}_{\mathrm{i}}=\frac{2\left(7.10 \mathrm{m}\right)}{\left(2.6 \mathrm{s}\right)} \\ =\frac{14.2 \mathrm{m}}{3.2 \mathrm{s}} \\ =5.46 \mathrm{m}/\mathrm{s} \end{gathered}$$

 

Thus, the box has not been pushed with the same initial speed in each case.

Step 5: (b) Determination of the ratio between greatest and the smallest initial speed

From the above answer, the greatest and the smallest initial speed are 6.72 m/s and 5.46 m/s respectively.

 

The ratio between the greatest and the smallest initial speed is expressed as:

 

  $\mathrm{r}=\frac{{\mathrm{v}}_1}{{\mathrm{v}}_2}$

Here, $\mathrm{r}$ is the ratio between the greatest and the smallest initial speed, ${\mathrm{v}}_1$ is the greatest and ${\mathrm{v}}_2$ is the smallest initial speed.

 

Substitute $6.72 \mathrm{m}/\mathrm{s}$ for ${\mathrm{v}}_1$ and $5.46 \mathrm{m}/\mathrm{s}$ for ${\mathrm{v}}_2$ in the above equation.

$$\begin{gathered} r=\frac{6.72 \mathrm{m}/\mathrm{s}}{5.46 \mathrm{m}/\mathrm{s}} \\ =1.2 \end{gathered}$$ 

 

Thus, the ratio between the greatest and the smallest initial speed is $1.2$.

Step 6: (c) Determination of the average horizontal force:

The equation of the average horizontal force is expressed as:

$\mathrm{F}=-\mathrm{ma}$ 

Here, $\mathrm{F}$ is the average horizontal force, $\mathrm{m}$ is the mass and $\mathrm{a}$ is the acceleration of the box.

 

In the first case, 

 

Substitute $8 \mathrm{kg}$ for $\mathrm{m}$ and $-2.1 \mathrm{m}/{\mathrm{s}}^2$ for $\mathrm{a}$ in the above equation.

 

 $$\begin{gathered} \mathrm{F}=-\left(8 \mathrm{kg}\right)\left(-2.1 \mathrm{m}/{\mathrm{s}}^2\right) \\ =16.8 \mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2 \\ =16.8 \mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2\times \frac{1 \mathrm{N}}{1\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2} \\ =16.8 \mathrm{N} \end{gathered}$$

 

In the second case, 

Substitute $11 \mathrm{kg}$ for $\mathrm{m}$ and $-2.1 \mathrm{m}/{\mathrm{s}}^2$ for $\mathrm{a}$ in the above equation.

 

 $$\begin{gathered} \mathrm{F}=-\left(11 \mathrm{kg}\right)\left(-2.1 \mathrm{m}/{\mathrm{s}}^2\right) \\ =23.1 \mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2 \\ =23.1 \mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2\times \frac{1 \mathrm{N}}{1\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2} \\ =23.1 \mathrm{N} \end{gathered}$$

 

In the third case, 

Substitute $15 \mathrm{kg}$ for m and $-2.1 \mathrm{m}/{\mathrm{s}}^2$ for $\mathrm{a}$ in the above equation.

 

 $$\begin{gathered} \mathrm{F}=-\left(15 \mathrm{kg}\right)\left(-2.1 \mathrm{m}/{\mathrm{s}}^2\right) \\ =31.5 \mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2 \\ =31.5 \mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2\times \frac{1 \mathrm{N}}{1\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2} \\ =31.5 \mathrm{N} \end{gathered}$$

 

In the fourth case, 

Substitute $20 \mathrm{kg}$ for m and $-2.1 \mathrm{m}/{\mathrm{s}}^2$ for a in the above equation.

 

 $$\begin{gathered} \mathrm{F}=-\left(20 \mathrm{kg}\right)\left(-2.1 \mathrm{m}/{\mathrm{s}}^2\right) \\ =42 \mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2 \\ =42 \mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2\times \frac{1 \mathrm{N}}{1\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2} \\ =42 \mathrm{N} \end{gathered}$$

Thus, the average horizontal force exerted by the floor on the box is not same.

Step 7: (c) Determination of the equation along with the graph

The graph total mass of the box and the magnitude of the total force has been drawn below:

                    

 

 In the above graph, it has been observed that with the increase in the mass, the force also increases.

The graph behaves like a straight line that has a slope of $2.1 \mathrm{m}/{\mathrm{s}}^2$ passing from the origin. Hence, the equation of the line can be expressed as:

 $\mathrm{F}=\mathrm{m}\left(2.1 \mathrm{m}/{\mathrm{s}}^2\right)$ 

               

 

Thus, the equation is  $\mathrm{F}=\mathrm{m}\left(2.1 \mathrm{m}/{\mathrm{s}}^2\right)$.