JEE MAIN - Physics (2023 - 1st February Morning Shift - No. 14)
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, $$g=10 \mathrm{~ms}^{-2}$$):
0.25
0.75
0.60
0.50
Explanation
$$
\begin{aligned}
& S=u t+\frac{1}{2} a t^2 \\\\
& 50=0+\frac{1}{2} \times a \times 100 \\\\
& a=1 \mathrm{~m} / \mathrm{s}^2 \\\\
& F-\mu m g=m a \\\\
& 30-\mu \times 50=5 \times 1 \\\\
& 50 \mu=25 \\\\
& \mu=\frac{1}{2}
\end{aligned}
$$
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