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Question
Consider the situation shown in the following figure. The system is released from rest and the block of mass 1 kg 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.
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Solution
\[\text{Given}, \]
\[ \text{m}_1 = 4 \text{ kg, m}_2 = 1 \text{kg } , \]
\[ \text{v}_2 = 0 . 3 \text{ m/s}\]
\[ \text{v}_1 = 2 \times 0 . 3 = 0 . 6 \text{ m/s}\]
\[\left( \text{ v}_1 = 2 \text{ v}_2 \text{ in this system } \right)\]
\[\text{ Height descended by the 1 kg block, h = 1 m } \]
\[\text{ Distance travelled by the 4 kg block } , \]
\[\text{ s } = 2 \times 1 = 2 \text{ m }\]
\[\text{ Initially the system is at rest . So, u }= 0\]
\[\text{ Applying work energy theoremwhich says that } \]
\[\text{ change in K . E . = Work done } \left( \text{ for the system } \right)\]
\[\left( \frac{1}{2} \right) \text{m}_1 \nu_1^2 + \left( \frac{1}{2} \right) \text{m}_2 \nu_2^2 = \left( - \mu \text{ R } \right) \text{ s } + \text{m}_2 \text{gh}\]
\[\frac{1}{2} \times 4 \times \left( 0 . 36 \right) + \frac{1}{2} \times 1 \times \left( 0 . 09 \right) [\text{ As, R = 4g = 40 N} ]\]
\[ = - \mu 40 \times 2 + 1 \times 40 \times 1\]
\[ \Rightarrow 0 . 72 + 0 . 045 = - 80 \mu + 10\]
\[ \Rightarrow \mu = \frac{9 . 235}{80} = 0 . 12\]
So, the coefficient of kinetic friction between the block and the table is 0.12 .
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