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QJEEM23Phc23c: JEE Main 2023 Physics - Class 11 JEEM23Phc23c Solution: A block of mass is placed at rest on a table of rough surface.…

A block of mass 5 kg5 \mathrm{~kg} is placed at rest on a table of rough surface. Now, if a force of 30 N30 \mathrm{~N} is applied in the direction parallel to surface of the table, the block slides through a distance of 50 m50 \mathrm{~m} in an interval of time 10 s10 \mathrm{~s} . Coefficient of kinetic friction is (given, g=10 ms−2g=10 \mathrm{~ms}^{-2} ):

Multiple Choice Options

  • A 0.25
  • B 0.75
  • C 0.6
  • D 0.5

Step-by-Step Verified Solution

amp;S=ut+12at2amp;50=0+12×a×100amp;a=1 m/s2amp;F−μmg=maamp;30−μ×50=5×1amp;50μ=25amp;μ=12\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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