Question

In Fig 8.54, a block slides along a track from one level to a higher level after passing through an intermediate valley. The track is frictionless until the block reaches the higher level. There a frictional force stops the block in a distance d. The block’s initial speed${\mathbf{v}}_{\mathbf{0}}\mathbf{}\mathbf{is}\mathbf{}\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$ , the height difference his 1.1 m, and${\mathit{\mu }}_{\mathbf{k}}\mathbf{}\mathit{i}\mathit{s}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{60}$ . Find d.

Short answer

The distance travelled by the block at the higher level is 1.2 m

Step-by-step solution

Listing the given quantities

The initial speed of the block = 6.0 m/s

The height difference between levels d = 1.1 m

The coefficient of friction is 0.60

Understanding the concept of conservation of energy

The block moving with initial speed travels past a frictionless intermediate valley. Then it rises to a higher level on the other side. Thus initial kinetic energy is converted to kinetic and potential energy. At this level, further track offers friction and kinetic energy of the block is lost as thermal energy and the block stops after travelling some distance. So, we apply the law of conservation of energy at each stage of motion.

Formula:

$$\begin{gathered} \mathrm{K}.\mathrm{E}=\frac{1}{2}{\mathrm{mv}}^2 \\ \mathrm{P}.\mathrm{E}=\mathrm{mgh} \\ {\mathrm{E}}_{\mathrm{th}}={\mathrm{F}}_{\mathrm{f}}\mathrm{d} \end{gathered}$$

Calculation of distance travelled by the block at the higher level

Using the law of conservation of energy for the first frictionless part of motion, we determine the speed of the block as it reaches higher level

$$\begin{gathered} K{E}_i=P{E}_f+K{E}_f \\ K{E}_F=K{E}_I-P{E}_F \\ \frac{1}{2}m{v}_f^2=\frac{1}{2}m{v}_f^2=mgh \\ \frac{1}{2}{v}_f^2=\frac{1}{2}\times 6.0^2-9.8\times 1.1 \\ {v}_f^2=2\times \left(18-10.8\right) \\ =14.4 \\ {v}_f=3.8\mathrm{m}/\mathrm{s} \end{gathered}$$

Now, this kinetic energy is lost as thermal energy. Hence we write

$$\begin{gathered} E_{th}=K{E}_{F_{}} \\ {\mu }_{K}mgd=\frac{1}{2}m{v}_f^2 \\ \mathrm{d}=\frac{v_f^2}{2{\mathrm{\mu }}_{\mathrm{k}}\mathrm{g}} \\ =\frac{14.4}{2\times 0.60\times 9.8} \\ =1.2\mathrm{m} \end{gathered}$$

Hence, the distance travelled by the block at the higher level is 1.2 m