A body of 10, kg is acted by a force of 129.4, N if g = 9.8,m/{sec ^2} . acceleration of block is 10

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4-2.Friction

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A body of $10\, kg$ is acted by a force of $129.4\, N$ if $g = 9.8\,m/{\sec ^2}$. The acceleration of the block is $10\,m/{s^2}$. What is the coefficient of kinetic friction

A

$0.03$

B

$0.01$

C

$0.3$

D

$0.25$

Solution

(c) Net force on the body = Applied force -Friction

$ma = F – {\mu _k}mg$

${\mu _k} = \frac{{F – ma}}{{mg}} = \frac{{129.4 – 10 \times 10}}{{10 \times 9.8}} = 0.3$

Standard 11

Physics

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