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JEE Mains · Physics · STD 11 - 4.2 friction

A block of mass \(5\,kg\) is placed at rest on a table of rough surface. Now, if a force of \(30\,N\) 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 \(10\,s\). Coefficient of kinetic friction is (given, \(g =10\,ms ^{-2}\))

  1. A \(0.60\)
  2. B \(0.75\)
  3. C \(0.50\)
  4. D \(0.25\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(0.50\)

Step-by-step Solution

Detailed explanation

\(S=u t+\frac{1}{2} a t^2\) \(50=0+\frac{1}{2} \times a \times 100\) \(a=1 m / s ^2\) \(F-\mu m g=m a\) \(30-\mu \times 50=5 \times 1\) \(50 \mu=25\) \(\mu=\frac{1}{2}\)
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