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A 30kg block is to be moved up an inclined plane at an anglue 30^(@) to the horizontal with a velocity of 5ms^(-1). If the frictional force retarding the motion is 150 N find the horizontal force required to move the block up the plane.(g=10 ms^(-2)). |
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Answer» Solution :The force REQUIRED to a body up an INCLINED plane is `F=mg sin theta+` frictional force `= 30(10)sin 30^(@)+150=300N`. If P is the horizontal force, `F=P COS theta` `P=(F)/(cos theta)=(300)/(cos theta)=(300xx2)/(sqrt(3))` `= 200sqrt(3)=346N`. |
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