Question

In Fig. 6-27, a box of Cheerios $\mathbf{(}\mathbf{mass}\mathbf{}{\mathbf{m}}_{\mathbf{c}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{kg}\mathbf{)}$and a box of Wheaties$\mathbf{(}\mathbf{mass}\mathbf{}{\mathbf{m}}_{\mathbf{w}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{kg}\mathbf{)}$are accelerated across a horizontal surface by a horizontal force $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{F}}$applied to the Cheerios box. The magnitude of the frictional force on the Cheerios box is $\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{N}$, and the magnitude of the frictional force on the Wheaties box is $\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{N}$. If the magnitude of$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{F}}$ is $\mathbf{12}\mathbf{}\mathbf{N}$, what is the magnitude of the force on the Wheaties box from the Cheerios box?

Short answer

The magnitude of the force on the Wheaties box from the Cheerios box is ${}^{8.5\mathrm{N}}$

Step-by-step solution

Given

Mass of Cheerios box,${}^{m_C=1.0kg}$

Mass of Wheaties box,${}^{m_W=3.0kg}$

External force apply on the Cheerios box,localid="1654578189411" ${}^{\stackrel{\rightharpoonup }{F}=12N}$

Frictional force on the Cheerios box, ${}^{{f_s}^C=2.0N}$

Frictional force on the Wheaties box, ${}^{{f_S}^W=1.0N}$

Determining the concept

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. Thus, using the free body diagram, the magnitude of the force on the Wheaties box from the Cheerios box can be found. First, find the acceleration of the system by using Newton’s 2nd law of motion. For this, consider both boxes as a one system, then by considering separately, find the contact force between them.

Formula:

$F_{net}=\sum ma$

where, F is the net force, m is mass and a is an acceleration.

Determining the free body diagram

Free Body Diagram of the Block:

Determining the magnitude of the force on the Wheaties box from the Cheerios box

By using Newton’s 2nd law of motion along horizontal direction to the (Cheerios+Wheaties Box),

$$\begin{gathered} F-f_s=m{a}_x \\ 12-6.0=4.0a_x \\ {a}_x=1.5m/s^2 \end{gathered}$$

The acceleration of the system is$1.5m/s^2$.

With this acceleration, apply the Newton’s 2nd law to the Wheaties box.Considering the F’ is the force applied by the cheerios box on the Wheaties box,

$$\begin{gathered} {\mathrm{F}}^{\text{'}}-4.0=(3.0)(1.5) \\ {\mathrm{F}}^{\text{'}}=8.5\mathrm{N} \end{gathered}$$

Therefore, the magnitude of the force on the Wheaties box from the Cheerios box is $8.5\mathrm{N}$.