91影视

Mass of the trunk is,$\mathrm{m}=50\mathrm{kg}$

Distance covered by the trunk at the incline is,$\mathrm{x}=6.0\mathrm{m}$

The angle of inclination is,$\mathrm{胃}=30掳$

The coefficient of friction is,$\mathrm{渭}=0.20$

We draw the free body diagram for the trunk. Using this, we can find the net force acting on the trunk. As we know the displacement of the trunk and the force are calculated, and we can find the work done by the applied force. Using the work done on a system by external force we can find the increase in the thermal energy.

Formulae:

The frictional force acting on a body, ${\mathrm{F}}_{\mathrm{f}}={\mathrm{渭}}_{\mathrm{s}}{\mathrm{F}}_{\mathrm{N}}$ (1)

The normal force acting on a body, ${\mathrm{F}}_{\mathrm{N}}=\mathrm{mg}$ (2)

The work done by an applied force, $\mathrm{W}=\mathrm{Force}脳\mathrm{displacement}$ (3)

The force due to Newton鈥檚 second law, $\mathrm{F}=\mathrm{ma}$ (4)

Free body diagram,

From the free body diagram, we can say that,

${\mathrm{F}}_{\mathrm{N}}={\mathrm{F}}_1\mathrm{蝉颈苍胃}+\mathrm{尘驳肠辞蝉胃}..................\left(5\right)$

And

${\mathrm{F}}_1\mathrm{肠辞蝉胃}-{\mathrm{F}}_{\mathrm{f}}-\mathrm{mg蝉颈苍胃}=\mathrm{ma}$

As the trunk is moving with constant velocity, a = 0 thus, the above equation using equations (1), (2), and (4) becomes,

$$\begin{gathered} {\mathrm{F}}_1\mathrm{肠辞蝉胃}={\mathrm{F}}_{\mathrm{f}}+\mathrm{mg蝉颈苍胃} \\ {\mathrm{F}}_1\mathrm{肠辞蝉胃}=\left({\mathrm{渭}}_{\mathrm{s}}+{\mathrm{F}}_{\mathrm{N}}\right)+\mathrm{mg蝉颈苍胃} \\ {\mathrm{F}}_1\mathrm{肠辞蝉胃}-\mathrm{mg蝉颈苍胃}=\left({\mathrm{渭}}_{\mathrm{s}}+{\mathrm{F}}_{\mathrm{N}}\right) \\ \frac{\left({\mathrm{F}}_1\mathrm{肠辞蝉胃}-\mathrm{mg蝉颈苍胃}\right)}{{\mathrm{渭}}_{\mathrm{s}}}={\mathrm{F}}_{\mathrm{N}} \\ {\mathrm{F}}_{\mathrm{N}}=\frac{\left({\mathrm{F}}_1\mathrm{肠辞蝉胃}-\mathrm{mg蝉颈苍胃}\right)}{{\mathrm{渭}}_{\mathrm{s}}} \end{gathered}$$

But, using equation (5), the above equation giving the horizontal force can be written as:

role="math" localid="1661486810675" $$\begin{gathered} \frac{\left({\mathrm{F}}_1\mathrm{肠辞蝉胃}-\mathrm{mg蝉颈苍胃}\right)}{{\mathrm{渭}}_{\mathrm{s}}}={\mathrm{F}}_1\mathrm{蝉颈苍胃}+\mathrm{尘驳肠辞蝉胃} \\ \left({\mathrm{F}}_1\mathrm{肠辞蝉胃}-\mathrm{mg蝉颈苍胃}\right)={\mathrm{渭}}_{\mathrm{s}}脳{\mathrm{F}}_1\mathrm{蝉颈苍胃}+\mathrm{尘驳肠辞蝉胃} \\ {\mathrm{F}}_1\mathrm{肠辞蝉胃}-{\mathrm{渭}}_{\mathrm{s}}{\mathrm{F}}_1\mathrm{蝉颈苍胃}={\mathrm{尘驳渭}}_{\mathrm{s}}\mathrm{肠辞蝉胃}+\mathrm{mg蝉颈苍胃} \\ {\mathrm{F}}_1脳\left(\mathrm{肠辞蝉胃}-{\mathrm{渭}}_{\mathrm{s}}\mathrm{蝉颈苍胃}\right)=\left({\mathrm{尘驳渭}}_{\mathrm{s}}\mathrm{肠辞蝉胃}+\mathrm{mg蝉颈苍胃}\right) \\ {\mathrm{F}}_1=\frac{\left({\mathrm{尘驳渭}}_{\mathrm{s}}\mathrm{肠辞蝉胃}+\mathrm{mg蝉颈苍胃}\right)}{\left(\mathrm{肠辞蝉胃}-{\mathrm{渭}}_{\mathrm{s}}\mathrm{蝉颈苍胃}\right)} \\ {\mathrm{F}}_1=\frac{\left(50\mathrm{kg}脳9.8\mathrm{m}/{\mathrm{s}}^2脳0.20脳\cos 30掳+50\mathrm{kg}脳9.8\mathrm{m}/{\mathrm{s}}^2脳\sin 30掳\right)}{\left(\cos 30掳-0.2脳\sin 30掳\right)} \\ {\mathrm{F}}_1=434.1\mathrm{kg}路\mathrm{m}/{\mathrm{s}}^2\left(\frac{1\mathrm{N}}{1\mathrm{kg}路\mathrm{m}/{\mathrm{s}}^2}\right) \\ {\mathrm{F}}_1=434.1\mathrm{N} \end{gathered}$$

Now, the work done by this horizontal force using equation (3) is given as:

$$\begin{gathered} \mathrm{W}=\left(434.1\mathrm{N}脳6\mathrm{m}\right)\cos 30掳 \\ =2.2脳{10}^3\mathrm{N}路\mathrm{m}\left(\frac{1\mathrm{J}}{1\mathrm{N}路\mathrm{m}}\right) \\ =2.2脳{10}^3\mathrm{J} \end{gathered}$$

Hence, the value of the work done is $2.2脳{10}^3\mathrm{J}$

Change in the P.E. of the trunk from the tree diagram is given as:

$$\begin{gathered} 鈭�\mathrm{P}.\mathrm{E}.=\mathrm{mgl蝉颈苍胃} \\ =\left(50\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(6\mathrm{m}\right)\sin 30掳 \\ =1.47脳{10}^3\mathrm{kg}路{\mathrm{m}}^2/{\mathrm{s}}^2\left(\frac{1\mathrm{J}}{1\mathrm{kg}路{\mathrm{m}}^2/{\mathrm{s}}^2}\right) \\ =1.47脳{10}^3\mathrm{J} \end{gathered}$$

Work done on a system by the external force is given by,

$\mathrm{W}=鈭�\mathrm{K}.\mathrm{E}.+鈭�\mathrm{P}.\mathrm{E}.+鈭�{\mathrm{E}}_{\mathrm{th}}$

In this case, the thermal energy can be calculated by substituting the values as:

$$\begin{gathered} \mathrm{W}=0+鈭�\mathrm{P}.\mathrm{E}.+鈭�{\mathrm{E}}_{\mathrm{th}} \\ 鈭�{\mathrm{E}}_{\mathrm{th}}=\mathrm{W}-鈭�\mathrm{P}.\mathrm{E}. \\ 鈭�{\mathrm{E}}_{\mathrm{th}}=2.24脳{10}^3\mathrm{J}-1.47脳{10}^3\mathrm{J} \\ 鈭�{\mathrm{E}}_{\mathrm{th}}=7.7脳{10}^2\mathrm{J} \end{gathered}$$

Hence, the value of the energy is $7.7脳{10}^2\mathrm{J}$