JEE Mains · Physics · STD 11 - 4.2 friction
A block of mass \(10\, kg\) starts sliding on a surface with an initial velocity of \(9.8\, ms ^{-1}\). The coefficient of friction between the surface and bock is \(0.5\). The distance covered by the block before coming to rest is: [use \(g =9.8\, ms ^{-2}\) ].........\(m\)
- A \(4.9\)
- B \(9.8\)
- C \(12.5\)
- D \(19.6\)
Answer & Solution
Correct Answer
(B) \(9.8\)
Step-by-step Solution
Detailed explanation
\(a =-\mu g =-0.5 \times 9.8=-4.9\) \(m / s ^{2}\) \(d =\frac{ v ^{2}}{2 a }=\frac{9.8 \times 9.8}{2(4.9)}\) \(=9.8\, m\)
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