In Fig. 8-55, a block slides along a path that is without friction until the block reaches the section of length L = 0.75 m, which begins at height h = 2.0 m on a ramp of angle $\mathbf{\theta }\mathbf{=}\mathbf{30}\mathbf{°}$ . In that section, the coefficient of kinetic friction is0.40. The block passes through point A with a speed of 8.0m/s. If the block can reach point B (where the friction ends), what is its speed there, and if it cannot, what is its greatest height above A?

The length of the section without friction is L = 0.75 m .

The height where the frictionless section begins h = 2.0 m

The angle of ramp is$\theta =30°$

The coefficient of kinetic friction is,${\mu }_k=0.40$

The speed of the block at point A is, ${\mathrm{v}}_{\mathrm{A}}=8.0\mathrm{m}/\mathrm{s}$.

First, find the speed ${\mathit{v}}_{\mathbf{c}}$at thefirst counter and the rough region as point C. By using this value of ${\mathit{v}}_{\mathbf{c}}$ , find thekinetic energy ${\mathit{k}}_{\mathbf{c}}$at point C. By using this value of ${\mathit{k}}_{\mathbf{c}}$ , find the distance dif d < L. Finally, using d = L, find the speed of the block at point B.

Formula:

The speed of the block at point C is,

$v_c=\sqrt{\left(v_A^2-2gh\right)}$

The kinetic energy is,

$K=\frac{1}{2}m{v}^2$

The equation for kinetic energy turns into thermal energy is,

$K_c=mgy+f_{k}d$

Let, first counter of the rough region be point C. The speed of the block at point C is given as

$$\begin{gathered} v_c=\sqrt{\left(v_A^2-2gh\right)} \\ {v}_c=\sqrt{\left[8.0-\left(2\times 9.8\times 2.0\right)\right]} \\ {v}_c=\sqrt{\left(24.8\right)} \\ {v}_c=4.980\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the kinetic energy at the beginning of its rough slide is given by

$$\begin{gathered} K_c=\frac{1}{2}m{v}_c^2 \\ {K}_c=\frac{1}{2}m\times 4.{980}^2 \\ {K}_c=12.4m \end{gathered}$$

m is the mass of the block.

Therefore, we get,

$$\begin{gathered} K_c=mgy+f_{K}d \\ 12.4m=mgd\sin \theta +{\mu }_{k}mgd\cos \theta \\ 12.4=gd\sin \theta +{\mu }_{k}gd\cos \theta \\ 12.4=9.8\times d\times \sin 30+0.40\times 9.8\times d\times \cos 30 \\ d=\frac{12.4}{\left(9.8\times d\times \sin 30+0.40\times 9.8\times d\times \cos 30\right)} \\ d=1.49m \end{gathered}$$

Here, we note that, d < L

Similarly, if d = L , we get,

$\frac{1}{2}m{v}^2=K_C-\left(mgL\sin \theta +{\mu }_{k}mgL\cos \theta \right)$

Therefore, the speedof the block at point B is

$$\begin{gathered} v_B=\sqrt{v_c^2-2gL\left(\sin \theta +{\mu }_k\cos \theta \right)} \\ {v}_B=\sqrt{4.{98}^2-2\times 9.8\times 0.75\times \left(\sin 30+0.40\times \cos 30\right)} \\ {v}_B=3.5\mathrm{m}/\mathrm{s} \end{gathered}$$

From the above calculations, we can say that block reaches point B.

Hence, the block can reach at point B and its speed $v_B$at point B is, $v_B=3.5\mathrm{m}/\mathrm{s}$