A block of mass $5 \mathrm{~kg}$ is placed at rest on a table of rough surface. Now, if a force of $30 \mathrm{~N}$ is applied in the direction parallel to surface of the table, the block slides through a distance of $50 \mathrm{~m}$ in an interval of time $10 \mathrm{~s}$. Coefficient of kinetic friction is (given, $g=10 \mathrm{~ms}^{-2}$):
Solution
$$
\begin{aligned}
& S=u t+\frac{1}{2} a t^2 \\\\
& 50=0+\frac{1}{2} \times a \times 100 \\\\
& a=1 \mathrm{~m} / \mathrm{s}^2 \\\\
& F-\mu m g=m a \\\\
& 30-\mu \times 50=5 \times 1 \\\\
& 50 \mu=25 \\\\
& \mu=\frac{1}{2}
\end{aligned}
$$
About this question
Subject: Physics · Chapter: Laws of Motion · Topic: Newton's First Law
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