Q36P - University Physics with Modern Physics

Question

 An advertisement claims that a particular automobile can “stop on a dime.” What net force would be necessary to stop $\mathbf{850}\mathbf{-}\mathbf{kg}$ automobile traveling initially at $\mathbf{45}\mathbf{.}\mathbf{0}\mathbf{km}\mathbf{/}\mathbf{h}$ in a distance equal to the diameter of a dime, $\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{cm}$?

Step-by-Step Solution

Answer

The net force required to stop the automobile is $3.6\times {10}^6\hspace{0.33em}\mathrm{N}$$3.6\times {10}^6\hspace{0.33em}\mathrm{N}$.

Step 1: Identification of the given data

The given data is listed below as:

• The mass of the automobile is, $m=850\mathrm{kg}$

• The speed of the automobile is, $\mathrm{v}=45.0 \mathrm{km}/\mathrm{h}\times \left(\frac{1000\mathrm{m}}{1\mathrm{km}}\right)\times \left(\frac{1\mathrm{h}}{3600\mathrm{s}}\right)=12.5\mathrm{m}/\mathrm{s}$

• The dime’s diameter is equal to the distance traveled by automobile $1.8\hspace{0.33em}"\mathrm{cm}"\times \left(\frac{{10}^{-2}\hspace{0.33em}\mathrm{m}}{ 1\mathrm{cm}}\right)=1.8\times {10}^{-2}\mathrm{m}$.

Step 2: Significance of the Newton’s second law

This law states that the force applied by an object is directly proportional to the product of the acceleration and the mass of the object. The acceleration occurs as a force acts on a particular mass.

Step 3: Determination of the net force

The equation of the acceleration of the automobile is expressed as:

$\begin{array}{l}{\mathrm{v}}^2={\mathrm{u}}^2+2\mathrm{as}\\ \mathrm{a}=\frac{{\mathrm{v}}^2-{\mathrm{u}}^2}{2\mathrm{s}}\end{array}$

                                                                                                                …(i)

Here, $\mathrm{V}$ is the final speed, $u$ is the initial speed, $a$ is the acceleration of the automobile, and $s$ is the distance traveled by automobile.

Use equation (i) for writing the equation of the net force.

$\begin{array}{l}\mathrm{F}=\mathrm{m}\\ ={\mathrm{m}}^2-{\mathrm{u}}^2\\ 2\mathrm{s}\end{array}$

Here, $m$ is the mass of the automobile.

Substitute all the values in the above equation.

$$\begin{gathered} F=\left(850\mathrm{kg}\right)\frac{{\left(12.5\mathrm{m}/\mathrm{s}\right)}^2}{2\left(1.8\times {10}^{-2}\right)} \\ =\left(850\mathrm{kg}\right)\frac{\left(156.25{\mathrm{m}}^2/{\mathrm{s}}^2\right)}{} \\ =\left(850\mathrm{kg}\right)\left(4340.27\mathrm{m}/{\mathrm{s}}^2\right) \\ =3.6\times {10}^6 \mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

$$\begin{gathered} F = \left(850\mathrm{kg}\right)\frac{{\left(12.5\mathrm{m}/\mathrm{s}\right)}^2-{\left(0\right)}^2}{2\left(1.8\times {10}^{-2}\mathrm{m}\right)} \\ = \left(850\mathrm{kg}\right)\frac{\left(156.25{\mathrm{m}}^2/{\mathrm{s}}^2\right)}{\left(0.036\mathrm{m}\right)} \\ = \left(850\mathrm{kg}\right)\left(4340.27\mathrm{m}/{\mathrm{s}}^2\right) \\ =3.6\times {10}^6\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Hence, further as:

$$\begin{gathered} F=3.6\times {10}^6 \mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\times \left(\frac{1\mathrm{N}}{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}\right) \\ =3.6\times {10}^6 \mathrm{N} \end{gathered}$$

$\mathrm{Thus}, \mathrm{the} \mathrm{net} \mathrm{force} \mathrm{required} \mathrm{to} \mathrm{stop} \mathrm{the} \mathrm{automobile} \mathrm{is} 3.6\times {10}^6\mathrm{N}.$