A block of 1 kg is stopped against a wall by applying a force F perpendicular to the wall. If $μ$ = 0.2 the minimum value of F will be:

980 N

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49N

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98 N

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490 N

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Solution

The correct option is B 49N
Given,

Weight = friction force

$m g = μ F$

$F = \frac{m g}{μ} = \frac{1 \times 9.8}{0.2} = 49 N$

Hence, required force is $49 N$

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