JEE Mains · Physics · STD 11 - 4.2 friction
A body of mass \(10\,kg\) is moving with an initial speed of \(20\,m / s\). The body stops after \(5\,s\) due to friction between body and the floor. The value of the coefficient of friction is (Take acceleration due to gravity \(g =10\; ms ^{-2}\))
- A \(0.2\)
- B \(0.3\)
- C \(0.5\)
- D \(0.4\)
Answer & Solution
Correct Answer
(D) \(0.4\)
Step-by-step Solution
Detailed explanation
\(a=-\mu g\) \(\because v=u+a t\) \(0=20+(-\mu \times 10) \times 5\) \(50 \mu=20\) \(\mu=\frac{2}{5}=0.4\)
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