A 10sqrt(3) 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))

A 10sqrt(3) kg box has to move up an inclined slope of 60^(@) to the h

1.	A 10sqrt(3) 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))
Answer» `300xx(2)/(sqrt(3))N` `300xx(sqrt(3))/(2)N` 300 N 150 N ANSWER :C