91影视

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 ${\mathrm{t}}_1=2.0\mathrm{s}$.

Newton鈥檚 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.

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

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

Here, localid="1664877856996" ${\mathrm{v}}_{\mathrm{f}}$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.

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

鈥�.. (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.

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

鈥�.. (2)

The above equation can also be written as:

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

鈥�.. (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.

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.

localid="1664879615026" $$\begin{gathered} {\mathrm{v}}_{\mathrm{i}}=\frac{2\left(8.22\mathrm{m}\right)}{\left(2.8\mathrm{s}\right)} \\ =\frac{16.44\mathrm{m}}{2.8\mathrm{s}} \\ =5.87\mathrm{m}/\mathrm{s} \end{gathered}$$

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.

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 $1.2$.

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.

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

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.

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

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.

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

In the fourth case,

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

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

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

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