A block of mass 2 kg slides down an incline plane of inclination $30 °$ . The coefficient of friction between block and plane is 0.5. The contact force between block and plank is :

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20 N

5 \sqrt{15} N

5 \sqrt{7} N

10 \sqrt{3} N

answer is D.

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Detailed Solution

Contact force

f = \sqrt{N^{2} + f_{R}^{2}}

$f = \sqrt{{(mg \cos θ)}^{2} + {(μ mg \cos θ)}^{2}}$

$f = mg \cos θ \sqrt{(1^{2} + μ^{2})}$

$f = 2 \times 10 \cos 30 \sqrt{1 + ({\frac{1}{2}}^{2})}$

$f = 20 \frac{\sqrt{3}}{2} \frac{\sqrt{5}}{2}$

$f = 5 \sqrt{3} \sqrt{5}$

$f = 5 \sqrt{15} N$

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