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A 30kg box has to move up an inclined pl...

A `30kg` box has to move up an inclined plane of slope `30^(@)` the horizontal with a unform velocity of `5 ms^(-1)`. If the frictional force retarding the motion is `150N`, the horizontal force required to move the box up is `(g =ms^(-2))` .

Text Solution

Verified by Experts


The force required to move a body up an inclined plane is
` F = "mg sin" theta + f_(k)`
`= 30(10) "sin" 30^(@) + 150 = 300 N`
If P is the horizontal force , F = P cos `theta`
`P = (F)/("cos" theta) = (300)/("cos" 30^(@))`
`= (300 xx 2)/(sqrt3) = 200 sqrt3 = 346 N`
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