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A 30 kg box has to move up an inclined slope of $30^{0}$ to the horizontal at a uniform velocity of 5 $m s^{- 1}$ . If the frictional force retarding the motion is 150N, the horizontal force required to move up is (g=10 $m s^{- 2}$ )

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detailed solution

Correct option is A

Fcos300 = mgsinθ + fFcos300 = mgsin300+ fF32=30×10×12+150=300 ⇒F=300×23N

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A 30 kg box has to move up an inclined slope of 300 to the horizontal at a uniform velocity of 5 ms-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 30 kg box has to move up an inclined slope of $30^{0}$ to the horizontal at a uniform velocity of 5 $m s^{- 1}$ . If the frictional force retarding the motion is 150N, the horizontal force required to move up is (g=10 $m s^{- 2}$ )