A block of mass 10 kg starts sliding on a surface with an initial velocity of 9.8 ms$-$1. The coefficient of friction between the surface and block is 0.5. The distance covered by the block before coming to rest is :
[use g = 9.8 ms$-$2]
Solution
<p>$S = {{{u^2}} \over {2a}} = {{{u^2}} \over {2(\mu g)}}$</p>
<p>$= {{{{(9.8)}^2}} \over {2 \times 0.5 \times (9.8)}}$</p>
<p>$= {{9.8} \over 1}$</p>
<p>$= 9.8$ m</p>
About this question
Subject: Physics · Chapter: Laws of Motion · Topic: Newton's First Law
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