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 block is 0 . 5 . The distance covered by the block before coming to rest is :[use g = 9 . 8 ms - 2 ]

Question

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 block is 0 . 5 . The distance covered by the block before coming to rest is :[use g = 9 . 8 ms - 2 ], 9 . 8 m, 4 . 9 m, 12 . 5 m, 19 . 6 m

MARKS App · Accepted Answer

Given initial velocity u = 9 . 8 m s - 1 . Acceleration a = μ g = 0 . 5 × 9 . 8 = 4 . 9 m s - 2 Using, v 2 = u 2 + 2 a s , here, final velocity v = 0 . So, 0 = 9 . 8 2 + 2 4 . 9 s ⇒ s = 9 . 8 m

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