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A body of mass 600 g slides on a rough horizontal surface. If the frictional force is 2.0 N.Then the magnitude of reaction is (Take $g = 10 m / s^{2}$ )

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$5 \sqrt{10} N$

5 N

7 N

$2 \sqrt{10} N$

answer is D.

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Detailed Solution

$m = 600 g = 6 \times 10^{- 1} kg$

$f = 2 N, R = ?$

$N = mg$

$= 6 \times 10^{- 1} \times 10 = 6 N$

We know

$Reaction force, R = \sqrt{N^{2} + f^{2}}$

$∴ R = \sqrt{36 + 4} = \sqrt{40}$

$∴ R = 2 \sqrt{10} N$

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