A block of mass 1 kg is held at rest against a rough vertical surface by pushing by a force F horizontally. The coefficient of friction is 0.5. When

F = 40 N, friction on the block is 20 N.

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F = 30 N, friction on the block is 10 N.

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F = 20 N, friction on the block is 10 N.

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Minimum value of force F to keep block at rest is 20 N.

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Solution

The correct option is C F = 20 N, friction on the block is 10 N.

$m = 1 k g$
$μ = 0 \cdot 5$
$f_{L} = μ N = μ F - (i)$
$f_{L} = 0.5 F$
$F_{moving} = m g$

$F_{m o v i n g} = 10 N$

$\Rightarrow f_{L} = μ F = 10 N$

$\Rightarrow F = \frac{f_{L}}{μ} = \frac{10}{0.5} = 20 N$

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