Q24E - University Physics with Modern Physics

Question

 A student of mass $\mathbf{45}\mathbf{ }\mathbf{kg}$ jumps off a high diving board. What is the acceleration of the earth toward her as she accelerates towards the earth with an acceleration of  $\mathbf{9}\mathbf{.}\mathbf{8}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$? Use  $\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{24}}$ for the mass of the earth, and assume that the net force on the earth is the force of gravity she exerts on it.

Step-by-Step Solution

Answer

The acceleration of the earth towards the student is $7.35\times {10}^{-23} \mathrm{m}/{\mathrm{s}}^2$.

Step 1: Identification of the given data

The given data can be listed below as,

• The mass of a student is, ${\mathrm{m}}_{\mathrm{s}}=45 \mathrm{kg}$.

• The acceleration of the student towards the earth is, ${\mathrm{a}}_{\mathrm{s}}=9.8 \mathrm{m}/{\mathrm{s}}^2$.

• The mass of the Earth is, ${\mathrm{m}}_{\mathrm{e}}=6.0\times {10}^{24} \mathrm{kg}$.

Step 2: Significance of linear acceleration.

Whenever an object of specific mass moves in a particular direction under the action of an external force, then there would be a linear acceleration exists. The concept of Newton’s law of motion helps to obtain the value of the acceleration.

Step 3: Determination of the force exerts by the earth on the student.

According to Newton’s second law of motion, the relation of force exerts by the earth on the student is expressed as,

$\mathrm{F}={\mathrm{m}}_{\mathrm{s}}{\mathrm{a}}_{\mathrm{s}}$

Here, $F$is the force exerts by the earth on the student.

Substitute $45\mathrm{kg}$ for $m_s$ and $9.8\mathrm{m}/{\mathrm{s}}^2$ for ${\mathrm{a}}_{\mathrm{s}}$ in the above equation.

$$\begin{gathered} F=\left(45 \mathrm{kg}\right)\left(9.8 \mathrm{m}/{\mathrm{s}}^2\right) \\ =\left(441 \mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\right)\times \left(\frac{1\mathrm{N}}{1 \mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}\right) \\ =441 \mathrm{N} \end{gathered}$$

Step 4: Determination of the acceleration of the earth towards the student.

The relation of acceleration of the earth towards the student is expressed as,

$\begin{array}{l}\mathrm{F}={\mathrm{m}}_{\mathrm{a}}^{\mathrm{t}}\\ 4_{\mathrm{z}}=\frac{\mathrm{F}}{{\mathrm{m}}_{\mathrm{m}}}\end{array}$

Here, ${\mathrm{a}}_{\mathrm{e}}$is the acceleration of the earth towards the student.

Substitute all the known values in the above equation.

Substitute $441\mathrm{N}$ for $F$ and $6.0\times {10}^{24}$ for $m_e$ in the above equation.

$$\begin{gathered} {\mathrm{a}}_{\mathrm{e}}=\left(\frac{441\mathrm{N}}{6.0\times {10}^{24}\mathrm{kg}}\right) \\ =7.35\times {10}^{-23} \mathrm{N}/\mathrm{kg} \\ =\left(7.35\times {10}^{-23} \mathrm{N}/\mathrm{kg}\right)\times \left(\frac{1\mathrm{m}/{\mathrm{s}}^2}{1\mathrm{N}/\mathrm{kg}}\right) \\ =7.35\times {10}^{-23} \mathrm{N}/\mathrm{kg} \end{gathered}$$

 

Thus, the acceleration of the earth towards the student is$7.35\times {10}^{-23} \mathrm{m}/{\mathrm{s}}^2$.