Laws of Motion

JEE Main level, single correct, medium

156 of 210

Question 156 of 210

A block of mass 10kg10kg starts sliding on a surface with an initial velocity of 9.8ms−19.8{ms}^{-1}. The coefficient of friction between the surface and block is 0.50.5. The distance covered by the block before coming to rest is :[use g=9.8ms−2g=9.8{ms}^{-2}]

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Answer: (A) 9.8m9.8m

Given initial velocity u=9.8ms−1u=9.8m{s}^{-1}.
Acceleration a=μg=0.5×9.8=4.9ms−2a=\mu g=0.5\times 9.8=4.9m{s}^{-2}
Using, v2=u2+2as{v}^{2}={u}^{2}+2as, here, final velocity v=0v=0.
So, 0=(9.8)2+2(4.9)s⇒s=9.8m0={\left(9.8\right)}^{2}+2\left(4.9\right)s\Rightarrow s=9.8m

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