A block of wood of mass 0.5kg is placed on a plane making angle $30^{0}$ with the horizontal. If the coefficient of friction between the surface of contact of the body and the plane is 0.2, what force parallel to the plane is required to keep the body sliding down with uniform velocity?

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1.6 N

3.2 N

6.4 N

4.8 N

answer is B.

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Detailed Solution

F = m g (\sin θ - μ_{k} \cos θ)

$= \frac{1}{2} \times 9.8 \times (\frac{1}{2} - \frac{1}{5} \frac{\sqrt{3}}{2}) = \frac{4.9}{2} (1 - \frac{\sqrt{3}}{5})$
$= \frac{4.9}{2} \times (\frac{5 - \sqrt{3}}{5}) = \frac{4.9}{2} \times (\frac{5 - 1.732}{5}) = 1.6 N$

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