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Q.

A 30kg box has to move up an inclined slope of 30^o to the horizontal at a uniform velocity of 5ms⁻¹. If the frictional force retarding the motion is 150N, the horizontal force required to move up is (g=10 ms⁻²)

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a

$300 \times \frac{2}{\sqrt{3}}$ N

b

$300 \times \frac{\sqrt{3}}{2}$ N

c

300 N

d

150 N

answer is A.

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Detailed Solution

Question Image

Force of friction = 150N

⟹Fcos30=mgsin30^∘+F

⟹F $\frac{\sqrt{3}}{2}$ =30x10x $\frac{1}{2}$ +150=300

⟹F = 300 x $\frac{2}{\sqrt{3}}$ N

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A 30kg box has to move up an inclined slope of 30o to the horizontal at a uniform velocity of 5ms−1. If the frictional force retarding the motion is 150N, the horizontal force required to move up is (g=10 ms−2)