91Ó°ÊÓ

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 ${\mathrm{μ}}_{\mathrm{k}}=0.25$
• The angle of horizontal with which the force is applied on the crate=$30°$

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

$$\begin{gathered} \mathbf{W}\mathbf{=}\stackrel{\mathbf{→}}{\mathbf{F}}\mathbf{·}\stackrel{\mathbf{→}}{\mathbf{s}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{Fs}\left(\mathrm{³¦´Ç²õÏ•}\right) \end{gathered}$$

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

(a)

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

$\mathrm{Fcos}30°>\mathrm{μ}\left(\mathrm{mg}+\mathrm{F}\mathrm{²õ¾\pm ²Ôθ}\right)$

Solve the above equation and find F.

$\mathrm{F}>\frac{\mathrm{μ³¾²\mu }}{\cos 30°-\mathrm{μ²õ¾\pm ²Ô}30°}$

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

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

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

(b)

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

$$\begin{gathered} \mathrm{W}=\stackrel{→}{\mathrm{F}}·\stackrel{→}{\mathrm{s}} \\ =\mathrm{Fs}\left(\mathrm{³¦´Ç²õÏ•}\right) \end{gathered}$$

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

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

role="math" localid="1664860011291" $$\begin{gathered} \mathrm{W}=\mathrm{Fs}\cos 30° \\ =95.8\mathrm{N}×4.5\mathrm{m}×\left(\frac{\sqrt{3}}{2}\right) \\ =373.4\mathrm{N}·\mathrm{m} \\ \mathrm{W}=373.4\mathrm{J} \end{gathered}$$

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

(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$-373.4\mathrm{J}$

(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.

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.