91影视

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

$W=鈭�k$

Work done is

$W=Fd\cos 胃$

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

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

$F_{netx}=0$

Therefore,

$$\begin{gathered} {\mathrm{F}}_{\mathrm{app}}-\mathrm{T}\mathrm{肠辞蝉胃}=0 \\ {\mathrm{F}}_{\mathrm{app}}=\mathrm{T}\mathrm{肠辞蝉胃} \end{gathered}$$

Also,

$F_{nety}=0$

Therefore,

$$\begin{gathered} -{\mathrm{F}}_{\mathrm{g}}+{\mathrm{T}}_{\sin }\mathrm{胃}=0 \\ {\mathrm{F}}_{\mathrm{g}}=\mathrm{T}\mathrm{蝉颈苍胃} \end{gathered}$$ 鈥�(颈颈)

Dividing equation (ii) by (i),

$$\begin{gathered} \frac{\mathrm{mg}}{{\mathrm{F}}_{\mathrm{app}}}=\mathrm{迟补苍胃} \\ {\mathrm{F}}_{\mathrm{app}}=\frac{\mathrm{mg}}{\mathrm{迟补苍胃}} \end{gathered}$$

Now, calculate the value of angle $胃$.

$$\begin{gathered} \mathrm{肠辞蝉胃}=\frac{\mathrm{d}}{\mathrm{L}} \\ \mathrm{胃}={\cos }^{-1\left(\frac{\mathrm{d}}{\mathrm{L}}\right)} \\ ={\cos }^{-1}\left(\frac{4.00\mathrm{m}}{12.0\mathrm{m}}\right) \\ =70.52掳 \end{gathered}$$

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

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

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,

$$\begin{gathered} \mathrm{h}=\mathrm{L}\mathrm{蝉颈苍胃} \\ =12\mathrm{m}脳\sin \left(70.5掳\right) \\ =12\mathrm{m}脳0.942 \\ =11.31\mathrm{m} \end{gathered}$$

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,

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

Using equation 3 and values of m and g,

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

Work done by the tension in the rope,

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

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

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,${\mathrm{w}}_{\mathrm{app}=}{\mathrm{F}}_{\mathrm{app}}\mathrm{x}.$ .