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A body of mass 400 g slides on a rough horizontal surface. If the frictional force is 3 N, then the magnitude of contact force is (take $g = 10 {ms}^{- 2})$

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4 N

3 N

7 N

5 N

answer is D.

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Detailed Solution

$\begin{array}{l} m = 400 g = 0 .4 kg \\ f_{s} = 3 N \\ N = mg = 0 .4 \times 10 = 4 N \\ R = \sqrt{N^{2} + f^{2}} = \sqrt{4^{2} + 3^{2}} \\ = \sqrt{9 + 16} = \sqrt{25} \\ R = 5 N \end{array}$

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