Q70P - University Physics with Modern Physics

Question

A 5.00 - kg box sits at rest at the bottom of a ramp that is 8.00 m long and is inclined at $\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{°}$ above the horizontal. The coefficient of kinetic friction is ${\mathit{\mu }}_{\mathbf{k}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{40}$, and the coefficient of static friction is ${\mathit{\mu }}_{\mathbf{s}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{43}$. What constant force F , applied parallel to the surface of the ramp, is required to push the box to the top of the ramp in a time of 6.00 s ?

Step-by-Step Solution

Answer

The constant force applied parallel to the surface is 43.56 N .

Step 1: Identification of the given data

The given data can be listed below as:

• The mass of the box is 5.00 kg.
• The length of the ramp is = 8.00 m .
• The angle of inclination of the box is $\theta =30.0°$.
• The coefficient of the kinetic friction is ${\mathrm{\mu }}_{\mathrm{k}}=0.40$.
• The coefficient of the static friction is ${\mathrm{\mu }}_{\mathrm{s}}=0.43$.
• The time required for the box to cover the distance is t = 6.00 s .

Step 2: Significance of the friction

Friction is described as the force that mainly resists the relative motion of an object on any surface. The frictional force is equal to the product of the frictional coefficient and the normal reaction force.

Step 3: Determination of the constant force

The free body diagram of the box has been drawn below:

                     

In the above figure, the frictional force f is acting. The weight mg that is the product of the mass of the box and acceleration due to gravity is being directed downwards. The weight has been divided into two components such as mgsin$30°$ and mgcos $30°$ respectively. 

 

The equation of the acceleration of the box is expressed as:

 

$$\begin{gathered} \mathrm{d}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2 \\ \mathrm{d}-\mathrm{ut}=\frac{1}{2}{\mathrm{at}}^2 \\ 2\left(\mathrm{d}-\mathrm{ut}\right)={\mathrm{at}}^2 \\ \mathrm{a}=\frac{2\left(\mathrm{d}-\mathrm{ut}\right)}{{\mathrm{t}}^2} \end{gathered}$$  

 

Here, d is the distance moved by the box that is the length of the ramp, u is the initial speed of the box,  is the time required for the box to cover the distance and  is the acceleration of the box.

 

As the box was at rest initially, then the initial speed of the box is zero.

 

Substitute the values in the above equation.

 

$$\begin{gathered} \mathrm{a}=\frac{2\left(8.00\mathrm{m}-\left(0\right)\mathrm{t}\right)}{{\left(6.00 \mathrm{s}\right)}^2} \\ =\frac{\left(16.00\mathrm{m}\right)}{\left(36.00 {\mathrm{s}}^2\right)} \\ =0.44 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$ 

 

The equation of the net force is expressed as:

 

F = ma                                                                                                                          …(i)

 

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

 

According to the free body diagram, the equation of the force is expressed as:

 

$\mathrm{F}=\mathrm{f}-\mathrm{mg} \sin 30°$                                                                                                          …(ii)

 

Here, f is the frictional force.

 

Equalling the equation (i) and (ii)

 

 $\mathrm{F}=\mathrm{f}+\mathrm{mgsin}30°+\mathrm{ma}$                                                                                                …(iii)

 

The equation of the frictional force is expressed as:

 

$F={\mu }_{k}mg\cos \theta$  

 

Here, ${\mathrm{\mu }}_{\mathrm{k}}$ is the coefficient of the kinetic friction and   is the angle between the ground and the box.

 

Substitute the above value in the equation (iii).

 

$\mathrm{F}={\mathrm{\mu }}_{\mathrm{k}}\mathrm{mgcos\theta }+\mathrm{mgsin}30°+\mathrm{ma}$ 

 

Substitute all the values in the above equation.

 

$$\begin{gathered} \mathrm{F}=\left(\begin{array}{c}\left(0.40\right)\left(5.00\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\cos 30.0°+\left(5.00\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\sin 30.0°\\ +\left(5.00\mathrm{kg}\right)\left(0.44\mathrm{m}/{\mathrm{s}}^2\right) \end{array}\right) \\ =\left(19.6\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\right)0.86+\left(49\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\right)\left(0.5\right)+2.2\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2 \\ =16.86\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2+24.5\mathrm{kg}.\mathrm{m}/\mathrm{s}+2.2\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2 \\ =43.56\mathrm{N} \end{gathered}$$ 

 

Thus, the constant force applied parallel to the surface is 43.56 N .