Gujarati
4-2.Friction
easy

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

A

$0.5$

B

$0.2$

C

$0.4$

D

$0.8$

Solution

(c) ${\mu _k} = \frac{F}{R} = $$\frac{{19.6}}{{5 \times 9.8}} = \frac{2}{5} = 0.4$ 

Standard 11
Physics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.