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

A body of mass $4 kg$ is at rest on a horizontal table. The coefficient of friction between body and table is 0.3 . Find the acceleration of the body if a horizontal force applied on the body is 20 N

\[(a)\;2.06 m/s^2 \quad (b)\;2.6 m/s^2 \quad (c)\;1.82 m/s^2 \quad(d)\;1.06 m/s^2 \]
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1 Answer

$R=mg$
$\quad=4 \times 9.8$
$\quad=39.2 N$
Therefore $ f=\mu mg$
$\quad \qquad= 0.3 \times 39.2N$
$\quad\qquad=11.76 N$
$F-f\; =ma$
$20-11.76=4a$
$\qquad \quad 8.24=4a$
$\qquad\qquad a= 2.06 m/s^2$
Hence a is the correct answer.

 

answered Jul 5, 2013 by meena.p
edited Jan 26, 2014 by meena.p
 

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