Question

In Fig. 8-67, a small block is sent through point A with a speed of 7 m/sIts path is without friction until it reaches the section of length L = 12 m, where the coefficient of kinetic friction is 0.70. The indicated heights are${\mathbf{h}}_{\mathbf{1}}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}$and${\mathit{h}}_{\mathbf{2}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}$What are the speeds of the block at (a) point B and (b) point C? (c) Does the block reach point D? If so, what is its speed there; if not, how far through the section of friction does it travel?

Short answer

a) The speed of the ball at point B is 12.9 m/s

b) The speed of the ball at point C is 11.3 m/s

c) The block does not reach point D.

Step-by-step solution

The given data

The initial speed of the block is,$v_i=7m/s$

The length of the section is,$L=12m$

The coefficient of the kinetic friction is, $\mu =0.70$

The indicated heights are, $h_1=6m$and $h_2=2m$.

Understanding the concept of energy

We use the principle of conservation of energy at point A and Point B. We get the equation in terms of kinetic energy and potential energy. From this, we can find the speed of the block at point B. Using the same procedure at point B and Point C, we can find the speed of the block at point C.

Formulae:

The kinetic energy of the body in motion,$KE=\frac{1}{2}m{v}^2$ (1)

The potential energy of a body at a height,$PE=mgh$ (2)

The frictional acceleration of the body,$a=-\mu g$ (3)

The third equation of kinematic motion,$v_f^2=v_i^2+2ax$ (4)

a) Calculation of the speed of the ball at point B

From the figure and for conservation of energy and using the equations (1) and (2), the speed of the ball at B is given as:

$$\begin{gathered} \left(\begin{array}{c}totalenergyat\\ pointA\end{array}\right)=\left(\begin{array}{c}totalenergyat\\ pointB\end{array}\right) \\ \frac{1}{2}m{v}_i^2+mg{h}_1=\frac{1}{2}m{v}_f^2 \\ m\times \left(\frac{1}{2}{v}_i^2+g{h}_1\right)=\frac{1}{2}m{v}_f^2 \\ \left(\frac{1}{2}{v}_i^2+g{h}_1\right)=\frac{1}{2}{v}_f^2 \\ 2\times \left(\frac{1}{2}{v}_i^2+g{h}_1\right)=\frac{1}{2}{v}_f^2 \\ {v}_f=\sqrt{v_i^2+2g{h}_1} \\ =\sqrt{{\left(7m/s\right)}^2+2\times 9.8m/s^2\times 6.0m} \\ =\sqrt{166.6m^2/s^2} \\ =12.9m/s \end{gathered}$$

Hence the value of the speed is 12.9 m/s

b) Calculation of the speed of the ball at point C

From the figure and for conservation of energy and using the equations (1) and (2), the speed of the ball at C is given as:

$$\begin{gathered} \left(\begin{array}{c}totalenergyat\\ pointB\end{array}\right)=\left(\begin{array}{c}totalenergyat\\ pointC\end{array}\right) \\ \frac{1}{2}m{v}_f^2=\frac{1}{2}m{v}_i^2+mg{h}_2 \\ \frac{1}{2}{v}_i^2=\left(\frac{1}{2}{v}_f^2+g{h}_2\right) \\ {v}_f=\sqrt{v_i^2+2g{h}_2} \\ =\sqrt{{\left(12.9m/s\right)}^2+2\times 9.8m/s^2\times 20m} \\ =\sqrt{166.6-39.2}m/s \\ =\sqrt{127.4}m/s \\ =11.3m/s \end{gathered}$$

Hence the value of the speed is 11.3 m/s

c) Calculation of the speed of the ball at point D

Acceleration on the rough patch can be given using equation (3) as:

$$\begin{gathered} a=-0.70\times 9.8m/s^2 \\ =-6.86m/s^2 \end{gathered}$$

We can find the distance traveled by the block on the rough patch before it comes to a stop using the kinematic equation (4) as follows:

$$\begin{gathered} {\left(0m/s\right)}^2={\left(11.3m/s\right)}^2-2\times 6.8m/s^2\times X \\ {\left(11.3m/s\right)}^2=2\times 6.8m/s^2\times X \\ \frac{127.7m^2/s^2}{2\times 6.8m/s^2}=x \\ x=9.31m \end{gathered}$$

Now the distance traveled by the block on the rough patch is 9.31 m but the actual length of the path is 12 m .

From this, we can conclude that the block will not reach point D.

It will travel the distance 9.31 m and at this point, the speed of the block will be zero.