logo

Ask Questions, Get Answers

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

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

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

1 Answer

$\mu_k=\large\frac{F}{R}$
$\qquad=\large\frac{19.6}{5 \times 9.8}$
$\qquad=\large\frac{2}{5}$
$\qquad=0.4$
Hence b is the correct answer. 

 

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

Related questions

...