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A block of mass $0.1 \mathrm{~kg}$ is held against a wall by applying a horizontal force of $5 \mathrm{~N}$ on it. If $\mu_{s}$ between the wall and the block is $0.5$, the magnitude of the frictional force acting on the block is
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The correct answer is:
$0.98 \mathrm{~N}$
Since, the block is at rest, the upward frictional force is equal to the weight or gravitational pull of the Earth i.e.,
$$
F_{S}=w=m g=0.1 \times 9.8=0.98 \mathrm{~N}
$$
$$
F_{S}=w=m g=0.1 \times 9.8=0.98 \mathrm{~N}
$$
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