Question

A box of bananas weighing 40.0 N rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.40 , and the coefficient of kinetic friction is 0.20 . (a) If no horizontal force is applied to the box and the box is at rest, how large is the friction force exerted on it? (b) What is the magnitude of the friction force if a monkey applies a horizontal force of 6.0 N to the box and the box is initially at rest? (c) What minimum horizontal force must the monkey apply to start the box in motion? (d) What minimum horizontal force must the monkey apply to keep the box moving at constant velocity once it has been started? (e) If the monkey applies a horizontal force of 18.0 N , what is the magnitude of the friction force and what is the box’s acceleration?

Short answer

1. The frictional force exerted on the box is, 0 N .
2. The magnitude of the frictional force is, 6 N .
3. The minimum horizontal force to start the box is, 16 N.
4. The minimum horizontal box to keep the box moving is, 8 N .
5. The magnitude of the frictional force is, 8 N and acceleration of box is, $2.45\mathrm{m}/{\mathrm{s}}^2$

Step-by-step solution

Identification of the given data:

The given data can be listed as below.

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

magnitude of the frictional force:

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.

(a) Determination of friction force exerted on the box when there is no horizontal force.

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.

(b) Determination of magnitude of friction force when horizontal force is 6.0 N.

The magnitude of the friction force is expressed as,

$f_s={\mu }_s\mathrm{N}$

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

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

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

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 .

(c) Determination of minimum horizontal force:

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

$f_s={\mu }_s\mathrm{N}$

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

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

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

Hence, required minimum horizontal force is 16 N .

(d) Determination of minimum horizontal force apply to keep box moving:

The expression for frictional force is expressed as,

$f_k={\mu }_{k}N$

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

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

role="math" localid="1667644025239" $$\begin{gathered} f_s=0.20\times 40.0N \\ =8\hspace{0.17em}N \end{gathered}$$

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

(e) Determination of magnitude of frictional force and acceleration of the box:

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

$f_k={\mu }_{k}N$

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

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

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.

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

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