Only two horizontal forces act on a 3.0 kg body that can move over a frictionless floor. one force is 9.0 n, acting due east, and the other is 9.2 n, acting 58° north of west. what is the magnitude of the body's acceleration?
The first thing you should do for this case is to find the horizontal and vertical components of the forces acting on the body. We have then: Horizontal = 9-9.2cos (58) = 4.124742769 N. Vertical = 9.2sin (58) = 7.802042485 N Then, the resulting net force is: F = √ ((4.124742769) ^ 2 + (7.802042485) ^ 2) = 8.825268826 N Then by definition: F = m * a Clearing the acceleration: a = F / m a = (8.825268826) / (3.0) = 2.941756275 m / s ^ 2 answer: The magnitude of the body's acceleration is 2.941756275 m / s ^ 2
