A block of mass $10 \,kg$. moving with acceleration $2 \,m / s ^2$ on horizontal rough surface is shown in figure The value of coefficient of kinetic friction is ...........
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(a)
$a=2\,m / s ^2$
$g=9.8\,m / s ^2$
$Friction force = \mu g$
$=\mu \times 10 \times 9.8$
$=98\,\mu$
Net force $=$ External force $=$ friction force
$m \times a =40-98\,\mu$
$2 \times 10 =40-98\,\mu$
$98 \mu =40-20$
$=20$
$\mu=\frac{20}{98}=\frac{10}{49}=0.20$
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