Q4E - University Physics with Modern Physics

Question

Suppose the worker in Exercise 6.3 pushes downward at an angle of  $\mathbf{30}\mathbf{°}$ below the horizontal (a) What magnitude of force must the worker apply to move the crate at constant velocity? (b) How much work is done on the crate by this force when the crate is pushed at a distance of 4.5 m? (c) How much work is done on the crate by friction during this displacement? (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) $\mathrm{F}=95.8 \mathrm{N}$

(b) $\mathrm{W}=379.26 \mathrm{J}$

(c) ${\mathrm{W}}_{\mathrm{f}}=-379.26 \mathrm{J}$

 

(d) ${\mathrm{W}}_{\mathrm{n}}=0 \mathrm{J}$ and ${\mathrm{W}}_9=0 \mathrm{J}$

 

(e) ${\mathrm{W}}_{\mathrm{net}}=0$

Step 1: Identification of the given data

The given data is listed below as-

• The distance covered by the crate when the crate is pushed is,  $\mathrm{s}=4.5 \mathrm{m}$
• The mass of the crate is,  $\mathrm{m}=30 \mathrm{kg}$ 
• The coefficient of friction between the crate and the floor is  ${\mathrm{\mu }}_{\mathrm{k}} =0.25$
• The angle of horizontal with which the force is applied on the crate= $30°$

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 

 $$\begin{gathered} \mathbf{W}\mathbf{=}\overrightarrow{\mathbf{F}}\mathbf{·}\overrightarrow{\mathbf{s}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{Fs}\left(\mathrm{cos\varphi }\right) \end{gathered}$$

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 to move the crate at constant velocity

(a)

To move the crate at a constant velocity, the following condition should be followed-

 $\mathrm{Fcos}30°>\mathrm{\mu }\left(\mathrm{mg}+\mathrm{F} \mathrm{sin\theta }\right)$ 

Solve the above equation and find F.

 $\mathrm{F}>\frac{\mathrm{\mu mg}}{\cos 30°-\mathrm{\mu sin}30°}$ 

Here, $\mathrm{\mu }$ 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  ${\mathrm{\mu }}_{\mathrm{k}} =0.25 , \mathrm{m}=30 \mathrm{kg} \mathrm{and} \mathrm{g}=10 \mathrm{m}·{\mathrm{s}}^{-2}$

 $$\begin{gathered} \mathrm{F}>\frac{0.25\times 30 \mathrm{kg}\times 10 \mathrm{m}·{\mathrm{s}}^{-2}}{{\displaystyle \frac{\sqrt{3}}{2}}-\left(0.25{\displaystyle \frac{1}{3}}\right)} \\ \mathrm{F}=95.8 \mathrm{N} \end{gathered}$$

  

Thus, the magnitude of force applied by the worker to move the crate at constant velocity is  $\mathrm{F}=95.8 \mathrm{N}$.

Step 4: Determination of work done on the crate when the crate is pushed at a distance of 4.5 m

(b)

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

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

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

Work done on the crate by the worker’s force $\mathrm{F}=98 \mathrm{N}$ will be

$$\begin{gathered} \mathrm{W}=\mathrm{Fs} \cos 30° \\ =95.8 \mathrm{N}\times 4.5 \mathrm{m}\times \left(\frac{\sqrt{3}}{2}\right) \\ =373.4 \mathrm{N}·\mathrm{m} \\ \mathrm{W}=373.4 \mathrm{J} \end{gathered}$$ 

 

Thus, work done on the crate when the crate is pushed at a distance of 4.5 m is  $\mathrm{W}=373.4 \mathrm{J}$.

Step 5: Determination of work done on the crate by friction during displacement

(c)

The work done on the crate due to friction is opposite to the force applied by the factory worker.

Therefore.

 $$\begin{gathered} {\mathrm{W}}_{\mathrm{f}}=-\mathrm{W} \\ =-373.4 \mathrm{J} \end{gathered}$$

Thus, work done on the crate by friction during displacement is  $-373.4 \mathrm{J}$

Step 6: Determination of work done on the crate by the normal force and by gravity

(d)

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

 $$\begin{gathered} {\mathrm{W}}_{\mathrm{n}}=\mathrm{Fs} \cos 90° \\ =0 \mathrm{J} \end{gathered}$$

Work done on the crate by the gravity will be

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

Here,   and s are perpendicular to each other.

Thus, work done on the crate by the normal force and by gravity is 0 J and 0 J respectively.

Step 7: Determination of total work done on the crate (e)

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

 $$\begin{gathered} {\mathrm{W}}_{\mathrm{net}}=\mathrm{W}+{\mathrm{W}}_{\mathrm{f}}+{\mathrm{W}}_{\mathrm{n}}+{\mathrm{W}}_{\mathrm{g}} \\ =373.4 \mathrm{J}-373.4 \mathrm{J}+0 \mathrm{J}-0 \mathrm{J} \\ =0 \mathrm{J} \end{gathered}$$

 

Thus, the total work done on the crate is 0 J.