91影视

The law of conservation of energy states that the total energy of an isolated system remains constant.

Formulae:

The distance of the projectile motion,$\mathrm{h}=\mathrm{诲蝉颈苍胃}$ (1)

The potential energy of the body, PE = mgh (2)

Applying the law of conservation of energy,

(3)

The increase in the thermal energy due to friction, $鈭�{\mathrm{E}}_{\mathrm{th}}={\mathrm{渭贵}}_{\mathrm{r}}$ (4)

If the plane is frictionless, then using equations (1), (2), and (3), we can get that distance traveled by the block is given as:

$$\begin{gathered} {\mathrm{KE}}_{\mathrm{i}}+{\mathrm{PE}}_{\mathrm{i}}={\mathrm{KE}}_{\mathrm{f}}+{\mathrm{PE}}_{\mathrm{f}} \\ \left(\frac{1}{2}\right){\mathrm{mv}}_0^2+0\mathrm{J}+\mathrm{mgdsin}\left(\mathrm{胃}\right) \\ \mathrm{d}=\frac{{\mathrm{v}}_0^2}{2\mathrm{gsin}\left(\mathrm{胃}\right)} \\ \mathrm{d}=\frac{{\left(5\mathrm{m}/\mathrm{s}\right)}^2}{2\mathrm{x}9.8\mathrm{m}/{\mathrm{s}}^2\mathrm{xsin}\left(30掳\right)} \\ \mathrm{d}=2.55\mathrm{m} \end{gathered}$$

Hence, the distance value is 2.55 m.

Now, if friction is available,$\mathrm{渭}=0.40$

Using equations(1), (2), and (3),we can get that the distance traveled by plane if the plane has friction can be given as:

$$\begin{gathered} {\mathrm{KE}}_{\mathrm{i}}+{\mathrm{PE}}_{\mathrm{i}}={\mathrm{KE}}_{\mathrm{f}}+{\mathrm{PE}}_{\mathrm{f}} \\ \left(\frac{1}{2}\right){\mathrm{mv}}_0^2=\mathrm{mgdsin}\left(\mathrm{胃}\right)+\mathrm{渭尘驳诲辞蝉}\left(\mathrm{胃}\right) \\ \mathrm{d}=\frac{{\mathrm{v}}_0^2}{2g(\sin \left(\mathrm{胃}\right)+\mathrm{渭}\cos \left(\mathrm{胃})\right)} \\ \mathrm{d}=\frac{{\left(5\mathrm{m}/\mathrm{s}\right)}^2}{2\mathrm{x}9.8\mathrm{m}/{\mathrm{s}}^{2}x(\sin \left(30掳\right)+0.4\cos \left(30掳)\right)} \\ \mathrm{d}=1.51\mathrm{m} \end{gathered}$$

Hence, the distance value is 1.51 m.

Thermal energy is generated by friction. Thus, it is given using equation (4) as follows:

$$\begin{gathered} {\mathrm{E}}_{\mathrm{thermal}}=\mathrm{渭尘驳诲肠辞蝉}\left(\mathrm{胃}\right) \\ =0.4\mathrm{x}5\mathrm{kgx}9.8\mathrm{m}/{\mathrm{s}}^2\mathrm{x}1.51\mathrm{m}\mathrm{x}\cos \left(30掳\right) \\ =25.6\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2\left(\frac{1\mathrm{J}}{1\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2}\right) \\ =25.6\mathrm{J} \end{gathered}$$

Hence, the increase in thermal energy is 25.6 J.

If the block slides down against frictional force, the speed of the block at the bottom using equations (1), (2), and (3) is given as:

$$\begin{gathered} {\mathrm{KE}}_{\mathrm{top}}+{\mathrm{PE}}_{\mathrm{top}}={\mathrm{KE}}_{\mathrm{bottom}}+{\mathrm{PE}}_{\mathrm{bottom}}+\mathrm{work}\mathrm{done} \\ 0+\mathrm{mgdsin}\left(\mathrm{胃}\right)=\frac{1}{2}{\mathrm{mv}}_{\mathrm{bottom}}^2+25.6\mathrm{J} \\ 5\mathrm{kg}\mathrm{x}9.8\mathrm{m}/{\mathrm{s}}^2脳1.51\mathrm{m}脳\sin \left(30掳\right)=\frac{1}{2}脳5.0\mathrm{kg}脳{\mathrm{v}}_{\mathrm{bottom}}^2+25.6\mathrm{J} \\ 37.0\mathrm{J}=\frac{1}{2}脳5.0\mathrm{kg}脳{\mathrm{v}}_{\mathrm{bottom}}^2+25.6\mathrm{J} \\ 11.4\mathrm{J}=\frac{1}{2}脳5.0\mathrm{kg}脳{\mathrm{v}}_{\mathrm{bottom}}^2 \\ {\mathrm{v}}_{\mathrm{bottom}}^2=4.56{\mathrm{m}}^2/{\mathrm{s}}^2 \\ {\mathrm{v}}_{\mathrm{bottom}}^2=2.14\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the speed of the block is 2.14 m/s.