A block of mass 10   kg starts sliding on a surface with an initial velocity of 9 . 8   ms - 1 .…

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]
  1. 9.8 m
  2. 4.9 m
  3. 12.5 m
  4. 19.6 m

Solution

Given initial velocity u=9.8 m s-1.

Acceleration a=μg=0.5×9.8=4.9 m s-2

Using, v2=u2+2as, here, final velocity v=0.

So, 0=9.82+24.9s⇒s=9.8 m

Asked in: JEE Main 2022 (24 Jun Shift 1)

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