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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Laws of Motion

A body of mass 2 kg is moving on the ground comes to rest after some time. The coefficient of kinetic friction between body and the ground is 0.2. The retardation in the body is

\[(a)\;9.8\; m/s^2 \quad (b)\;4.73\; m/s^2 \quad (c)\;2.16\; m/s^2 \quad (d)\;1.96 \;m/s^2 \]

1 Answer

We know that
$a=\mu g$
$\quad=0.2 \times 9.8$
$\quad=1.96 m/s^2$
Hence d is the correct answer. 

 

answered Jul 16, 2013 by meena.p
edited Feb 10, 2014 by meena.p
 

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