Q23E - University Physics with Modern Physics

Question

Boxes A and B are in contact on a horizontal, frictionless surface (Fig. E4.23). Box A has mass $\mathbf{20}\mathbf{ }\mathbf{kg}$ and box B has mass$\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{kg}$. A horizontal force of $\mathbf{250}\mathbf{N}$ is exerted on box A. What is the magnitude of the force that box A exerts on box B?

Step-by-Step Solution

Answer

The magnitude of the force that box A exerts on box B is $50 \mathrm{N}$.

Step 1: Identification of the given data

The given data can be listed below as,

• The mass of box A is, ${\mathrm{m}}_{\mathrm{A}}=20.0 \mathrm{kg}$.

• The mass of box B is, ${\mathrm{m}}_{\mathrm{B}}=5.0 \mathrm{kg}$.

• The magnitude of a horizontal force is, $\mathrm{F}=250\hspace{0.33em}\mathrm{N}$.

Step 2: Significance of Net Acceleration of Two Connected Objects.

When two objects are connected on a smooth horizontal surface (frictionless), then due to the applied horizontal force, the objects start moving forward with a specific acceleration.

Step 3: Determination the acceleration of the box system.

According to Newton’s second law of motion, the relation of acceleration of the box system is expressed as,

$\begin{array}{l}\mathrm{F}=\left({\mathrm{m}}_{\mathrm{A}}+\mathrm{m})\right)\\ \mathrm{a}=\frac{\mathrm{F}}{\left({\mathrm{m}}_{\mathrm{A}}+{\mathrm{m}}_8\right)}\end{array}$

Here, $a$is the acceleration of the box system.

Substitute $250\mathrm{N}$ for $F$, $20\mathrm{kg}$ for $m_A$ and $5\mathrm{kg}$ for $m_B$ in the above equation.

$$\begin{gathered} \mathrm{a}=\frac{250 \mathrm{N}}{\left(20 \mathrm{kg} + 5 \mathrm{kg}\right)} \\ =\left(10 \mathrm{N}/\mathrm{kg}\right)\times \left(\frac{1\mathrm{m}/{\mathrm{s}}^2}{1\mathrm{N}/\mathrm{kg}}\right) \\ =10\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

$\mathrm{Thus}, \mathrm{the} \mathrm{magnitude} \mathrm{of} \mathrm{the} \mathrm{force} \mathrm{that} \mathrm{box} \mathrm{A} \mathrm{exerts} \mathrm{on} \mathrm{box} \mathrm{B} \mathrm{is} 50\mathrm{N}.$

Step 4: Determination the magnitude of the force that box A exerts on box B.

The relation of magnitude of the force that box A exerts on box B is expressed as,

${\mathrm{F}}_{\mathrm{A}}={\mathrm{m}}_{\mathrm{B}}\mathrm{a}$

Here, $F_A$is the magnitude of the force that box A exerts on box B.

Substitute all the known values in the above equation.

Substitute $5.0\mathrm{kg}$ for $m_8$, $10\mathrm{m}/{\mathrm{s}}^2$ for $a$ in the above equation.

$$\begin{gathered} F_{A }= \left(5.0 \mathrm{kg}\right)\left(10\mathrm{m}/{\mathrm{s}}^2\right) \\ =5.0 \mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2 \\ =\left(5.0 \mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\right)\times \left(\frac{1\mathrm{N}}{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}\right) \\ =50 \mathrm{N} \end{gathered}$$

Thus, the magnitude of the force that box A exerts on box B is.