A 30 kg 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)).

A 30 kg block is to be moved up an inclined plane at an anglue 30^(@)

1.	A 30 kg 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)).
Answer» Solution :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)=(300xx2)/(sqrt(3))` `= 200 sqrt(3)=346 N`.