91影视

Mass of the steel safe$m=2200\mathrm{kg}$.

Speed of the steel safe before hitting the concrete$v=40\mathrm{m}/\mathrm{s}$.

The change in length of the steel safe$D/=0.6\mathrm{m}$.

A force is a push or pull on an object caused by the interaction of the thing with another object. Every time two things interact, a force is exerted on each of them.

The change in length of the safe is$\mathrm{D}1=0.06\mathrm{m}$. Using the approximation

role="math" localid="1656668384432" $\left(\begin{array}{c}{\stackrel{\mathrm{r}}{\mathrm{v}}}_{\mathrm{avg}}\end{array}鈮�\left(\begin{array}{c}{\stackrel{\mathrm{r}}{\mathrm{v}}}_{\mathrm{f}}+\end{array}{\stackrel{\mathrm{r}}{\mathrm{v}}}_{\mathrm{i}}\right)/2=\left(\mathrm{谩}0,0\tilde{\mathrm{n}}+\mathrm{谩}0,鈥�40,0\tilde{\mathrm{n}}\right)/2=\mathrm{谩},鈥�20,0\tilde{\mathrm{r}}\mathrm{m}/\mathrm{s}\right)$.

The time interval that the safe and the concrete were in contact is :

$$\begin{gathered} \mathrm{D}t=\frac{\mathrm{D}1}{\left|{{\displaystyle \stackrel{\mathrm{I}}{\mathrm{V}}}}_{\mathrm{avg}}\right|} \\ =\frac{0.06}{20} \\ =3\text{'}{10}^{鈥�3}\mathrm{s} \end{gathered}$$

Thus, the value of is $3\text{'}{10}^{鈥�3}\mathrm{s}$.

The force exerted is given by the formula$\stackrel{r}{F}=\frac{\mathrm{D}{\displaystyle \stackrel{\mathrm{I}}{\mathrm{p}}}}{\mathrm{D}t}$

Substituting the value of $\mathrm{D}t$as$3\text{'}{10}^{鈥�3}\mathrm{s}$and expanding the formula,

localid="1656669677695"

Thus, the force exerted by the safe on the concrete is .

The magnitude of gravitational force is:

$$\begin{gathered} \left|\stackrel{r}{F}\right|=\sqrt{{\left(0\right)}^2{\left(鈥�2.93\text{'}{10}^7\right)}^2+{\left(0\right)}^2} \\ =2.93\text{'}{10}^7\mathrm{N} \end{gathered}$$

The gravitational force on the safe is found using the formula${\mathit{F}}_{\mathbf{g}}\mathbf{=}\mathit{m}\mathit{g}$.

$$\begin{gathered} F_g=\left(2200\right)\left(9.8\right) \\ =2.16\text{'}{10}^4\mathrm{N} \end{gathered}$$

Thus, the gravitational force of the Earth on the safe is $2.16\text{'}{10}^4\mathrm{N}$.

Take the ratio of force exerted by the safe and thegravitational force of the Earth:

$$\begin{gathered} \frac{F_{impact}}{F_g}=\frac{2.93\text{'}{10}^7\mathrm{N}}{2.2\text{'}{10}^4\mathrm{N}} \\ =1.33\text{'}{10}^3 \end{gathered}$$

The force by the impact on the steel is $1.33\text{'}{10}^3$greater than the gravitational force on the safe, one factor that made the force is the time duration.