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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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A lift is moving downward with an acceleration equal to acceleration due to gravity. A body of mass M kept on the floor of the lift is pulled horizontally. If the coefficient of friction is $\mu$, then the frictional resistance offered by the body is

\[(a)\;Mg \quad (b)\;\mu Mg \quad (c)\;2 \mu Mg \quad(d)\;zero \]
Can you answer this question?
 
 

1 Answer

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For downward acceleration motion of lift
$R=m(g-0)$
Since $a=g;R=0$
Therefore $F=\mu R=0$
Hence d is the correct answer. 

 

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

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