Question

A 5.0 kgblock is projected at 5.0 m/sup a plane that is inclined at$\mathbf{30}\mathbf{°}$with the horizontal. How far up along the plane does the block go (a) if the plane is frictionless and (b) if the coefficient of kinetic friction between the block and the plane is 0.40? (c) In the latter case, what is the increase in thermal energy of the block and plane during the block’s ascent? (d) If the block then slides back down against the frictional force, what is the block’s speed when it reaches the original projection point?

Short answer

a) The block goes 2.55 m far up if the plane is frictionless.

b) The block goes 1.51 m far up if the plane has friction.

c) The increase in the thermal energy of the block and the plane during the block’s ascent is 25.6 J

d) The block’s speed when it reaches the original projection point is 2.14 m/s

Step-by-step solution

The given data

$$\begin{gathered} \mathrm{The}\mathrm{mass}\mathrm{of}\mathrm{the}\mathrm{block}\mathrm{is},\mathrm{m}=5.0\mathrm{kg} \\ \mathrm{The}\mathrm{angle}\mathrm{of}\mathrm{inclination}\mathrm{is},\mathrm{\theta }=30° \\ \mathrm{The}\mathrm{coefficient}\mathrm{of}\mathrm{kinetic}\mathrm{friction}\mathrm{is},\mathrm{\mu }=0.40 \\ \mathrm{The}\mathrm{initial}\mathrm{speed}\mathrm{of}\mathrm{the}\mathrm{block}\mathrm{is},{\mathrm{v}}_{\circ }=5.0\mathrm{m}/\mathrm{s} \\ \mathrm{The}\mathrm{final}\mathrm{speed}\mathrm{of}\mathrm{the}\mathrm{block}\mathrm{is},{\mathrm{v}}_{\mathrm{f}}=0\mathrm{m}/\mathrm{s} \end{gathered}$$

Understanding the concept of energy

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

Formulae:

The distance of the projectile motion,$\mathrm{h}=\mathrm{dsin\theta }$ (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, $∆{\mathrm{E}}_{\mathrm{th}}={\mathrm{\mu F}}_{\mathrm{r}}$ (4)

a) Calculation of the distance if the plane is frictionless

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

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

Hence, the distance value is 2.55 m.

b) Calculation of the distance if the plane has friction

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

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

$$\begin{gathered} {\mathrm{KE}}_{\mathrm{i}}+{\mathrm{PE}}_{\mathrm{i}}={\mathrm{KE}}_{\mathrm{f}}+{\mathrm{PE}}_{\mathrm{f}} \\ \left(\frac{1}{2}\right){\mathrm{mv}}_0^2=\mathrm{mgdsin}\left(\mathrm{\theta }\right)+\mathrm{\mu mgdos}\left(\mathrm{\theta }\right) \\ \mathrm{d}=\frac{{\mathrm{v}}_0^2}{2g(\sin \left(\mathrm{\theta }\right)+\mathrm{\mu }\cos \left(\mathrm{\theta })\right)} \\ \mathrm{d}=\frac{{\left(5\mathrm{m}/\mathrm{s}\right)}^2}{2\mathrm{x}9.8\mathrm{m}/{\mathrm{s}}^{2}x(\sin \left(30°\right)+0.4\cos \left(30°)\right)} \\ \mathrm{d}=1.51\mathrm{m} \end{gathered}$$

Hence, the distance value is 1.51 m.

c) Calculation of the increase of the thermal energy generated by the friction

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

$$\begin{gathered} {\mathrm{E}}_{\mathrm{thermal}}=\mathrm{\mu mgdcos}\left(\mathrm{\theta }\right) \\ =0.4\mathrm{x}5\mathrm{kgx}9.8\mathrm{m}/{\mathrm{s}}^2\mathrm{x}1.51\mathrm{m}\mathrm{x}\cos \left(30°\right) \\ =25.6\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2\left(\frac{1\mathrm{J}}{1\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2}\right) \\ =25.6\mathrm{J} \end{gathered}$$

Hence, the increase in thermal energy is 25.6 J.

d) Calculation of the speed of the block

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:

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

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