Question

Consider the situation shown in the figure (8-E2). The system is released from rest and the block of mass 1kg is found to have a speed 0⋅3 m/s after it has descended a distance of 1 m. Find the coefficient of kinetic friction between the block and the table.
Figure

Answer 1

$$\begin{gathered} \mathrm{Given}, \\ {m}_1=4\mathrm{kg},m_2=1\mathrm{kg}, \\ {v}_2=0.3\mathrm{m}/s \\ {v}_1=2\times 0.3=0.6\mathrm{m}/s \\ \left(v_1=2v_2\mathrm{in}\mathrm{this}\mathrm{system}\right) \\ \mathrm{Height}\mathrm{descen}\text{ded}\mathrm{by}\text{the}1\mathrm{kg}\mathrm{block},h=1\mathrm{m} \\ \mathrm{Distance}\mathrm{travelled}\mathrm{by}\text{the}4\mathrm{kg}\mathrm{block}, \\ s=2\times 1=2\mathrm{m} \\ \mathrm{Initially}\mathrm{the}\mathrm{system}\mathrm{is}\mathrm{at}\mathrm{rest}.\text{So, u}=0 \\ \mathrm{Applying}\mathrm{work}\mathrm{energy}\mathrm{theoremwhich}\mathrm{says}\mathrm{that} \\ \mathrm{change}\mathrm{in}\mathit{}K\mathit{.}E\mathit{.}=\text{W}\mathrm{ork}\mathrm{done}\left(\mathrm{for}\mathrm{the}\mathrm{system}\right) \\ \left(\frac{1}{2}\right)m_1{\nu }_1^2+\left(\frac{1}{2}\right)m_2{\mathrm{\nu }}_2^2=\left(-\mu R\right)s+m_{2}gh \end{gathered}$$
$$\begin{gathered} \frac{1}{2}\times 4\times \left(0.36\right)+\frac{1}{2}\times 1\times \left(0.09\right)[\mathrm{As},R=4g=40\mathrm{N}] \\ =-\mu 40\times 2+1\times 40\times 1 \\ \Rightarrow 0.72+0.045=-80\mu +10 \\ \Rightarrow \mu =\frac{9.235}{80}=0.12 \end{gathered}$$
So, the coefficient of kinetic friction between the block and the table is 0.12 .