Question

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?

Answer 1

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