Q3E - University Physics with Modern Physics

Question

Question: A factory worker pushes 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and the floor is 0.25. (a) What magnitude of force must the worker apply? (b) How much work is done on the crate by this force? (c) How much work is done on the crate by friction? (d) How much work is done on the crate by the normal force? (e) What is the total work done on the crate?

Step-by-Step Solution

Answer

 

(a) ${\mathrm{F}}_{\mathrm{worker}}=73.5\mathrm{N}$ 

(b) ${\mathrm{W}}_{\mathrm{worker}}=330.75 \mathrm{J}$  

(c) ${\mathrm{W}}_{\mathrm{k}}=-330.75 \mathrm{J}$  

 

(d) ${\mathrm{W}}_{\mathrm{n}}=0\hspace{0.33em}\mathrm{J}$and ${\mathrm{W}}_{\mathrm{g}}=0\hspace{0.33em}\mathrm{J}$ 

 

(e)  ${\mathrm{W}}_{\mathrm{net}}=0 \mathrm{J}$

Step 1: Identification of the given data

The given data is listed below as-

• The distance covered by crate along a horizontal level floor is, s=4.5 m  
• The mass of the crate is, m =30 kg   
• The coefficient of friction between the crate and the floor is ${\mathrm{\mu }}_{\mathrm{k}}=0.25$   

Step 2: Significance of the work done

Work done on a particle by a constant force $\overrightarrow{\mathbf{F}}$ during a linear displacement $\overrightarrow{\mathbf{s}}$ is given by 

$$\begin{gathered} \mathbf{W}\mathbf{=}\overrightarrow{\mathbf{F}}\mathbf{.}\overrightarrow{\mathbf{s}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{Fs}\mathbf{ }\mathbf{cos\varphi } \end{gathered}$$ 

If F and s are in the same direction then $\mathrm{\varphi }=0$ and if it is in opposite direction then $\mathrm{\varphi }=180°$.

Step 3: Determination of magnitude of force applied by the worker

(a)

The worker should apply a ${\mathrm{F}}_{\mathrm{worker}}$ force  that is equal to the kinetic friction force ${\mathrm{f}}_{\mathrm{k}}$ to overcome the coefficient of kinetic friction between the crate and the floor.

Therefore,

${\mathrm{F}}_{\mathrm{worker}}-{\mathrm{f}}_{\mathrm{k}}=0$  

The kinetic friction force has a constant magnitude 

$$\begin{gathered} {\mathrm{f}}_{\mathrm{k}}={\mathrm{\mu }}_{\mathrm{k}}\mathrm{n} \\ ={\mathrm{\mu }}_{\mathrm{k}}\mathrm{mg} \end{gathered}$$ 

Here, ${\mathrm{\mu }}_{\mathrm{k}}$ is the coefficient of friction between the crate and the floor, m is the mass of the crate and g is the gravitational constant.

For ${\mathrm{\mu }}_{\mathrm{k}}=0.25$, $\mathrm{m}=30\hspace{0.33em}\mathrm{kg}$ and $\mathrm{g}=9.8\hspace{0.33em}\mathrm{m}\cdot {\mathrm{s}}^{-2}$ 

Therefore, the kinetic friction force is given by

$$\begin{gathered} {\mathrm{f}}_{\mathrm{k}}=0.25\times 30\mathrm{kg}\times 9.8\mathrm{m}.{\mathrm{s}}^{-2} \\ =73.5 \mathrm{kg}.\mathrm{m}.{\mathrm{s}}^{-2} \\ =73.5 \mathrm{N} \end{gathered}$$  

Step 4: Determination of work done on the crate

(b)

The work done on a particle by constant force $\overrightarrow{F}$ during the displacement $\overrightarrow{s}$ is given by

$$\begin{gathered} \mathrm{W}=\overrightarrow{\mathrm{F}}.\overrightarrow{\mathrm{s}} \\ =\mathrm{F}\times \mathrm{s}\left(\mathrm{cos\varphi }\right) \end{gathered}$$ 

Here, $\varphi$ is the angle between F and s.

Work done on the crate by the worker’s force $F_{worker}$ push will be

$$\begin{gathered} {\mathrm{W}}_{\mathrm{worker}}={\mathrm{F}}_{\mathrm{worker}}\mathrm{s}\left(\cos 0\right) \\ ={\mathrm{F}}_{\mathrm{worker}}\mathrm{s} \\ =73.5 \mathrm{N}\times 4.5 \mathrm{m} \\ =330.75 \mathrm{N}.\mathrm{m} \end{gathered}$$  

 ${\mathrm{W}}_{\mathrm{worker}}=330.75 \mathrm{J}$

(c)

The kinetic force is opposite to the force applied by the worker.

The angle between kinetic friction force and displacement is 180⁰.

Therefore,

$$\begin{gathered} {\mathrm{W}}_{\mathrm{k}}={\mathrm{f}}_{\mathrm{k}}\mathrm{scos}180° \\ =-{\mathrm{f}}_{\mathrm{k}}\mathrm{s} \\ =-73.5 \mathrm{N}\times 4.5\mathrm{m} \\ =-330.75 \mathrm{J} \end{gathered}$$  

(d)

Work done on the crate by the normal force from the floor will be 

$$\begin{gathered} {\mathrm{W}}_{\mathrm{n}}={\mathrm{F}}_{\mathrm{worker}}\mathrm{scos}90° \\ =0 \mathrm{J} \end{gathered}$$  

Work done on the crate by the gravity will be

$$\begin{gathered} {\mathrm{W}}_{\mathrm{g}}={\mathrm{F}}_{\mathrm{s}} \cos \left(-90°\right) \\ =0 \mathrm{J} \end{gathered}$$ 

Here, ${\mathrm{F}}_{\mathrm{woker}}$ and s are perpendicular to each other.

(e)

Net work done on the crate will be the sum of all the individual work

$$\begin{gathered} {\mathrm{W}}_{\mathrm{net}}={\mathrm{W}}_{\mathrm{woker}}+{\mathrm{W}}_{\mathrm{k}}+{\mathrm{W}}_{\mathrm{n}}+{\mathrm{W}}_{\mathrm{g}} \\ =330.75 \mathrm{J}-330.75 \mathrm{J}+0\mathrm{J}-0\mathrm{J} \\ =0 \mathrm{J} \end{gathered}$$  

Thus, the magnitude of the force the worker must apply ${\mathrm{F}}_{\mathrm{woker}}=73.5 \mathrm{N}$ .

The various Work done on the crate by this force are ${\mathrm{W}}_{\mathrm{worker}}$, ${\mathrm{W}}_{\mathrm{k}}=-330.75\hspace{0.33em}\mathrm{J}$ , ${\mathrm{W}}_{\mathrm{n}}=0\hspace{0.33em}\mathrm{J}$ , ${\mathrm{W}}_{\mathrm{g}}=0\hspace{0.33em}\mathrm{J}$, and $W_{net}=0 J$ .