A body is moving along the horizontal surface with a velocity of 4 ~m / s . If the coefficient of kinetic friction is 0.2 , the distance travelled by body before coming to rest is (g=10 ~m / s^2 )

Correct answer: (C) 4 ~m. f=μ mg . ∴ a=-f / m=-(μ m g)/m=-μ g=-(0.2) 10=-2 v^2=u^2+2 a s s=4

Q116 / 150 · MHT CET 2020 (20 Oct Shift 2) · Physics

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 $\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)$

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

Hence, the correct answer is option: 'C'.