A block of mass 5 kg is placed at rest on a table of rough surface. Now, if a force of 30N is applied in the direction parallel to surface of the table, the block slides through a distance of 50 m in an interval of time 10s. Coefficient of kinetic friction is (given $g = 10 {ms}^{- 2}$ ):

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0.75

0.50

0.25

0.60

answer is A.

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Detailed Solution

$\begin{array}{l} \frac{1}{2} {at}^{2} = 50 \\ a = \frac{50 \times 2}{10^{2}} = 1 m / s^{2} \\ \frac{30 - μmg}{m} = 1 \\ 6 - 10 μ = 1 \\ μ = 0.5 \end{array}$

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