91Ó°ÊÓ

The given data can be listed as below.

• Weight of box of bananas is, W = 40 N
• The coefficient of kinetic friction is, ${\mathrm{μ}}_k=0.20$
• The coefficient of static friction is, ${\mathrm{μ}}_s=0.40$

Frictional force can be defined as that it is a force that resists the relative motion between the two surfaces when they are in contact with each other. The magnitude of frictional force basically depends on the roughness of the surfaces and the weight.

The expression for Newton’s third law can be expressed as,

${\mathrm{F}}_{AB}=-F_{\mathrm{BA}}$

Here, ${\mathrm{F}}_{AB}$Is force exerted on AB and, $-F_{BA}$ is equal and opposite force exerted on BA .

Hence, there is no horizontal force applied on banana box, the frictional force exerted on it 0 N according to above equation.

Hence, required frictional force is, 0 N.

The magnitude of the friction force is expressed as,

$f_s={\mathrm{μ}}_s\mathrm{N}$

Here, $f_s$ is the frictional force, ${\mathrm{μ}}_s$ is the coefficient of kinetic friction, N is the normal force.

Substitute 0.40 for ${\mathrm{μ}}_s$, 40.0 N for in the above equation.

$$\begin{gathered} f_s=0.40×40.0\mathrm{N} \\ =16\mathrm{N} \end{gathered}$$

Hence, applied force 6.0 N is less than $f_s$.

Magnitude of applied force is the applied force which is equal to $6.0N$.

Hence, magnitude of frictional force is 6.0 N .

The expression for the frictional force to start the motion is expressed as,

$f_s={\mathrm{μ}}_s\mathrm{N}$

Here, $f_s$is the frictional force, ${\mathrm{μ}}_s$ is the coefficient of static friction and N is the normal force on the box.

Substitute 0.40 for ${\mathrm{μ}}_s$, and 40.0 N for N in the above equation.

$$\begin{gathered} f_s=0.40×40.0N \\ =16N \end{gathered}$$

Hence, required minimum horizontal force is 16 N .

The expression for frictional force is expressed as,

$f_k={\mathrm{μ}}_{k}N$

Here, $f_k$is the frictional force due to motion, ${\mathrm{μ}}_k$ is the coefficient of kinetic friction and N is the normal force.

Substitute 0.40 for ${\mathrm{μ}}_s$, 40.0 N for N in the above equation.

role="math" localid="1667644025239" $$\begin{gathered} f_s=0.20×40.0N \\ =8 N \end{gathered}$$

Hence, required minimum horizontal force to keep box moving is 8 N .

The expression for the magnitude of frictional force can be expressed as,

$f_k={\mathrm{μ}}_{k}N$

Substitute role="math" localid="1667644081894" $0.20$ for ${\mathrm{μ}}_k$, 40.0 N for N in the above equation.

$$\begin{gathered} f_k=0.20×40.0N \\ =8N \end{gathered}$$

Magnitude of frictional force $f=f_k=8\mathrm{N}$.

Hence, the required magnitude of frictional force is 8 N .

The expression for the net force can be expressed as,

$F=F_a-f$

Here, $F_a$is horizontal force apply for acceleration, f is the magnitude of frictional force.

Substitute 18.0 N for $f_a$, and 8 N for f in the above equation.

$$\begin{gathered} \mathrm{F}=18.0\mathrm{N}-8\mathrm{N} \\ =10\mathrm{N} \end{gathered}$$

Hence, net force is 10 N .

The expression for the acceleration can be expressed as,

$$\begin{gathered} \mathrm{F}=\mathrm{m}\mathrm{a} \\ \mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}} \end{gathered}$$

Here, F is the net force and is the mass of the box.

Substitute 10.0 N for F , and $\frac{40.0\mathrm{N}}{9.8\mathrm{m}/{\mathrm{s}}^2}$ for m in the above equation.

$$\begin{gathered} \mathrm{a}=10\mathrm{N}×\frac{9.8\mathrm{m}/{\mathrm{s}}^2}{40.0\mathrm{N}} \\ =2.45\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Hence, the required acceleration is $2.45\mathrm{m}/{\mathrm{s}}^2$.