Question

Figure 5-23 shows the same breadbox in four situations where horizontal forces are applied. Rank the situations according to the magnitude of the box’s acceleration, greatest first.

Short answer

$\left(\mathrm{a}\right)>\left(\mathrm{b}\right)=\left(\mathrm{C}\right)=\left(\mathrm{d}\right)$

Step-by-step solution

Given information

Figure 5-23.

To understand the concept

The problem is based on Newton’s second law of motion. The law states that the acceleration of an object is dependent on the net force acting upon the object and the mass of the object. Acceleration of the box can be found by calculating the net force acting on each box. The net force would be the vector sum of all the forces acting on that box.

Formula:

${\mathbf{F}}_{\mathbf{net}}=\mathbf{ma}$

Step 3: To rank the situation according to the magnitude of the box’s acceleration, the greatest first

We assume that the right-hand side direction is positive.

So, for the figure (a),

$$\begin{gathered} F_a=6\mathrm{N}-3\mathrm{N} \\ =3\mathrm{N} \\ {\mathrm{a}}_{\mathrm{a}}=\frac{3\mathrm{N}}{\mathrm{m}} \end{gathered}$$

For figure (b),

$$\begin{gathered} F_b=60\mathrm{N}-58\mathrm{N} \\ =2\mathrm{N} \\ {\mathrm{a}}_{\mathrm{b}}=\frac{2\mathrm{N}}{\mathrm{m}} \end{gathered}$$

For figure (c),

$$\begin{gathered} {\mathrm{F}}_{\mathrm{c}}=15\mathrm{N}-13\mathrm{N} \\ =2\mathrm{N} \\ {\mathrm{a}}_{\mathrm{c}}=\frac{2\mathrm{N}}{\mathrm{m}} \end{gathered}$$

For figure (d),

role="math" localid="1656995312985" $$\begin{gathered} {\mathrm{F}}_{\mathrm{d}}=25\mathrm{N}+20\mathrm{N}-23\mathrm{N} \\ =2\mathrm{N} \\ {\mathrm{a}}_{\mathrm{d}}=\frac{2\mathrm{N}}{\mathrm{m}} \end{gathered}$$

From the above values of acceleration for each figure, we can conclude that,

$\left({\mathrm{a}}_{\mathrm{a}}\right)>\left({\mathrm{a}}_{\mathrm{b}}\right)=\left({\mathrm{a}}_{\mathrm{c}}\right)=\left({\mathrm{a}}_{\mathrm{d}}\right)$