A 230 kg crate hangs from the end of a rope of length L= 12.0 m . You push horizontally on the crate with a varying force $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{F}}$to move it distance d= 4.00 m to the side $\left(\mathrm{Fig}.7-44\right)$.(a) What is the magnitude of $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{F}}$when the crate is in this final position? During the crate’s displacement, what are (b) the total work done on it, (c) the work done by the gravitational force on the crate, and (d) the work $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{F}}$done by the pull on the crate from the rope? (e) Knowing that the crate is motionless before and after its displacement, use the answers to (b), (c), and (d) to find the work your force does on the crate. (f) Why is the work of your force not equal to the product of the horizontal displacement and the answer to (a)?

Mass of the crate is m = 230 g

The length of the rope is L = 12 m

Horizontal displacement is x = 4 m

The problem deals with the power and the work done. Work, energy, and power are the fundamental concepts of physics. Work is the displacement of an object when force is applied on it.

To solve this problem, use,

• x and y components of the force
• Work-energy theorem
• Work-force relation

Work done is

$W=∆k$

Work done is

$W=Fd\cos \theta$

Where, d is displacement, Wis the work done;F is force and $∆k$is change in kinetic energy.

As the system is in equilibrium, net force acting in x direction is zero,

$F_{netx}=0$

Therefore,

$$\begin{gathered} {\mathrm{F}}_{\mathrm{app}}-\mathrm{T}\mathrm{cos\theta }=0 \\ {\mathrm{F}}_{\mathrm{app}}=\mathrm{T}\mathrm{cos\theta } \end{gathered}$$

Also,

$F_{nety}=0$

Therefore,

$$\begin{gathered} -{\mathrm{F}}_{\mathrm{g}}+{\mathrm{T}}_{\sin }\mathrm{\theta }=0 \\ {\mathrm{F}}_{\mathrm{g}}=\mathrm{T}\mathrm{sin\theta } \end{gathered}$$ …(ii)

Dividing equation (ii) by (i),

$$\begin{gathered} \frac{\mathrm{mg}}{{\mathrm{F}}_{\mathrm{app}}}=\mathrm{tan\theta } \\ {\mathrm{F}}_{\mathrm{app}}=\frac{\mathrm{mg}}{\mathrm{tan\theta }} \end{gathered}$$

Now, calculate the value of angle $\theta$.

$$\begin{gathered} \mathrm{cos\theta }=\frac{\mathrm{d}}{\mathrm{L}} \\ \mathrm{\theta }={\cos }^{-1\left(\frac{\mathrm{d}}{\mathrm{L}}\right)} \\ ={\cos }^{-1}\left(\frac{4.00\mathrm{m}}{12.0\mathrm{m}}\right) \\ =70.52° \end{gathered}$$

Now, substitute the values and calculate ${\mathrm{F}}_{\mathrm{app}}$.

$$\begin{gathered} {\mathrm{F}}_{\mathrm{app}}=\frac{\mathrm{mg}}{\mathrm{tan\theta }} \\ =\frac{230\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2}{\tan \left(70.52\right)} \\ =797\mathrm{N} \end{gathered}$$

Hence, the force acting on the crate at the final position is$\left|\mathrm{F}\right|=797\mathrm{N}$

Initial velocity u = 0 and final velocity v= 0.

Hence, the initial kinetic energy

$$\begin{gathered} {\mathrm{K}}_{\mathrm{i}}=\frac{1}{2}{\mathrm{mv}}_{\mathrm{i}}^2 \\ =0 \end{gathered}$$

and the final kinetic energy is

$$\begin{gathered} {\mathrm{K}}_{\mathrm{i}}=\frac{1}{2}{\mathrm{mv}}_{\mathrm{i}}^2 \\ =0 \\ {\mathrm{W}}_{\mathrm{net}}={\mathrm{K}}_{\mathrm{f}}-{\mathrm{K}}_{\mathrm{i}} \\ =0\mathrm{J} \end{gathered}$$

Hence, the total work done is${\mathrm{W}}_{\mathrm{net}}=0\mathrm{J}$

From the figure,

$$\begin{gathered} \mathrm{h}=\mathrm{L}\mathrm{sin\theta } \\ =12\mathrm{m}\times \sin \left(70.5°\right) \\ =12\mathrm{m}\times 0.942 \\ =11.31\mathrm{m} \end{gathered}$$

Vertical displacement of crate,

$$\begin{gathered} \mathrm{y}=\mathrm{L}-\mathrm{h} \\ =12\mathrm{m}-11.311\mathrm{m} \\ =0.689\mathrm{m} \end{gathered}$$

Now,

$$\begin{gathered} {\mathrm{w}}_{\mathrm{g}}=\mathrm{m}\mathrm{g}\mathrm{y}\mathrm{cos\theta } \\ =\mathrm{m}\mathrm{g}\mathrm{y}\cos \left(180\right) \end{gathered}$$

Using equation 3 and values of m and g,

$$\begin{gathered} {\mathrm{w}}_{\mathrm{g}}=230\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2\times 0.689\times \left(-1\right) \\ =-1553.006\mathrm{J} \\ =1.55\mathrm{kJ} \end{gathered}$$

Work done by the tension in the rope,

$$\begin{gathered} {\mathrm{W}}_{\mathrm{T}}=\mathrm{Td}\mathrm{cos\theta } \\ =\mathrm{T}\mathrm{d}\cos 90 \\ =0 \end{gathered}$$

The angle between the tension in the rope and circular displacement d is $90°$.

Hence, the work done by the rope is${\mathrm{W}}_{\mathrm{T}}=0\mathrm{J}$

$$\begin{gathered} {\mathrm{W}}_{\mathrm{net}}={\mathrm{W}}_{\mathrm{T}}+{\mathrm{W}}_{\mathrm{app}}+{\mathrm{W}}_{\mathrm{g}} \\ 0=0+{\mathrm{W}}_{\mathrm{app}}-1553.006\mathrm{J} \\ {\mathrm{W}}_{\mathrm{app}}=1553.006\mathrm{J} \end{gathered}$$

Hence, thework done by the applied force is${\mathrm{W}}_{\mathrm{app}}=1.55\mathrm{kJ}$

During displacement, there are varying forces such as applied force, gravitational force, tension in the rope that are acting on the crate. Hence, consider the net force acting on the crate. Therefore, work done by the applied force cannot be the product of just ${\mathrm{F}}_{\mathrm{app}}$and displacement.

Hence, the work of the applied force is not equal to the product of the horizontal displacement and the applied force because during the displacement of the crate force acting on it is not the same. Hence,${\mathrm{w}}_{\mathrm{app}=}{\mathrm{F}}_{\mathrm{app}}\mathrm{x}.$ .