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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 :                                                                                                            

[used g = 9.8 ms-2]

Option 1 -

4.9 m

Option 2 -

9.8 m

Option 3 -

12.5 m 

Option 4 -

19.6 m

0 3 Views | Posted 2 months ago
Asked by Shiksha User

  • 1 Answer

  • V

    Answered by

    Vishal Baghel | Contributor-Level 10

    2 months ago
    Correct Option - 2


    Detailed Solution:

    a µg = 0.5 × 9.8 = 4.9 m/sec2

    u = 9.8 m/sec

    S =?

    v = 0

    v2 = u2 + 2as

      ⇒ 0 = 9 . 8 2 − 2 × 4 . 9 s

    ⇒ 0 = 9 . 8 × 9 . 8 − 9 . 8 s

    s = 9.8×9.89.8=9.8  m  

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