A body is moving along the horizontal surface with a velocity of $4 \mathrm{~m} / \mathrm{s}$. If the…

A body is moving along the horizontal surface with a velocity of $4 \mathrm{~m} / \mathrm{s}$. If the coefficient of kinetic friction is $0.2$, the distance travelled by body before coming to rest is $\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)$
  1. $8 \mathrm{~m}$
  2. $16 \mathrm{~m}$
  3. $4 \mathrm{~m}$
  4. $6 \mathrm{~m}$

Solution

$f=\mu \mathrm{mg}$. $\therefore a=-f / m=-\frac{\mu m g}{m}=-\mu g=-(0.2) 10=-2$ $v^{2}=u^{2}+2 a s$ $s=4$ ~

Asked in: MHT CET 2020 (20 Oct Shift 2)

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