Laws of Motion

JEE Main level, single correct, medium

194 of 210

Question 194 of 210

A marble block of mass 2 kg2 \mathrm{~kg} lying on ice when given a velocity of 6 m/s6 \mathrm{~m} / \mathrm{s} is stopped by friction in 10 s10 \mathrm{~s}. Then the coefficient of friction is

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Answer: (D) None of these

a=μg=610a=\mu g=\frac{6}{10} [ using v=u+at\mathrm{v}=\mathrm{u}+\mathrm{at} ]
⇒μ=610×g=610×10=0.06\Rightarrow \mu=\frac{6}{10 \times g}=\frac{6}{10 \times 10}=0.06

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