## Fundamentals of Physics Extended (10th Edition)

2.9 $m/s^2$
There are two forces acting on the body so first we need to find the net force. One force (9N) is acting due east and the other is $62^{\circ}$ North of West. We must first separate the second force into its x and y components, which we can do using trigonometry. The x component, Fx is given by Fx = Fcos(62) = 3.8N West and the vertical component, Fy is given by Fy = Fsin(62) = 7.1N North. We have two forces in the x direction so we need to find the net force, which is Fnet,x = 9N (East) - 3.8N (West) = 5.2N East. We then use the equation F = ma in the x direction to find $a_{x} = F_{x}/m =$ 5.2N/3kg = 1.7 $m/s^2$ We do the same in the y direction but there is only one force so $a_{y} = F_{y}/m =$ 7.1N/3kg = 2.4 $m/s^2$ To find the magnitude of a we need to take $\sqrt (a_{x} ^2 + a_{y}^2$ = $\sqrt (1.7^2 + 2.4^2)$ = 2.9 $m/s^2$