A block of mass $20 k g$ is pushed with a horizontal force of $90 N$ . If the coefficent of static & kinetic friction are $0.4$ & $0.3$ , the frictional force acting on the block is

90 N

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80 N

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60 N

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30 N

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Solution

The correct option is C $60 N$
Given that,

Mass of block, $M = 20 k g$

Weight (vertically downward), $W = 20 \times 10 = 200 N$

Normal reaction (vertically upward), $W = 200 N$

Now, static friction

$f_{s} = μ m g$

$f_{s} = 0.4 \times 200$

$f_{s} = 80 N$

So the block will move and kinetic friction would act.

Now, kinetic friction

$f_{k} = μ m g$

$f_{k} = 0.3 \times 200$

$f_{k} = 60 N$
As motion start then only kinetic frction will act.

Hence, the frictional force is $60 N$

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