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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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A block of mass $5 kg$ lies on a horizontal table. A force of $19.6 N$ is enough to keep the body sliding at uniform speed. The coefficient of sliding friction

\[(a)\;0.5 \quad (b)\;0.4 \quad (c)\;0.2 \quad (d)\;0.8\]

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$\qquad=\large\frac{19.6}{5 \times 9.8}$
Hence b is the correct answer. 


answered Jul 15, 2013 by meena.p
edited Feb 10, 2014 by meena.p

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