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 - <p>4.9 m</p>
Option 2 - <p>9.8 m</p>
Option 3 - <p>12.5 m&nbsp;</p>
Option 4 - <p>19.6 m</p>
3 Views|Posted 7 months ago
Asked by Shiksha User
1 Answer
V
7 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=9.8m  

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Physics Laws of Motion 2025

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