a 30kg box has to move up an inclined slope of 30° to horizontal at a uniform velocity of 5m/sec.If the frictional force retarding the motion is 150N the horizontal force to move up is
Answers
Answered by
6
F = m ( gsin30 + ugcos30)
= 30×(10×0.5 + 150)
= 30× (155)
= 4650
= 30×(10×0.5 + 150)
= 30× (155)
= 4650
Similar questions