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A 10sqrt3 kg box has to move up an incli...

A `10sqrt3` kg box has to move up an inclined slope of `60^(@)` to the horizontal at a uniform velocity of 5 `ms^(-1)` If the frictional force retarding the motion is 150N, the minimum force applied parallel to inclined plane to move up is (g=10 `ms^(-2)`)

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The force required to a body up an inclined plane is:
`F = mg sin theta +` frictional force.
`=30(10) sin 30^(@) + 150 = 300 N`
If P is the horizontal force, `F = P cos theta`
`P = F/(cos theta) = 300/(cos theta) = (300 xx 2)/sqrt(3) = 200sqrt(3) = 346 N`
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