A block of mass $5$ kg lies on a rough horizontal table. A force of $19.6\, N$ is enough to keep the body sliding at uniform velocity. The coefficient of sliding friction is
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(c) ${\mu _k} = \frac{F}{R} = $$\frac{{19.6}}{{5 \times 9.8}} = \frac{2}{5} = 0.4$
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