Question

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?

Short answer

The block can reach at point B and its speed ${\mathrm{v}}_{\mathrm{B}}$at point B is,${\mathrm{v}}_{\mathrm{B}}=3.5\mathrm{m}/\mathrm{s}$

Step-by-step solution

Given Data

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, ${\mathrm{v}}_{\mathrm{A}}=8.0\mathrm{m}/\mathrm{s}$.

Understand the concept

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,

$K_c=mgy+f_{k}d$

Find out if the block can reach point B (where the friction ends) and calculate the velocity of the block at that point

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.

Here, we note that if d < L , the kinetic energy turns entirely o thermal energy.

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, $v_B=3.5\mathrm{m}/\mathrm{s}$