"Only two forces act on an object (mass = 3.90 kg), as in the drawing. (F = 76.0 N.) Find the magnitude and direction (relative to the x axis) of the acceleration of the object. " (The angle is 45 degrees and the horizontal force is 40N)

Answer 1

Net force is F = 40 + 76•cosα =40 +76•0.707 = 93.74 N
a =F/m =93.74/3.9 = 24 m/s^2

Answer 2

I’ve found the figure to this problem, so the solution is following.
x- and y- components of net forces are

F1(x) = 40 N, F1(y) =0
F2 (x) =76•cosα = 93.74 N,
F2 (y)= 76•sin α = 93.74 N ,
Net force
F(x)=133.74 N,
F(y) = 93.74 N.
F = sqrt(133.74^2 + 93.74^2) = 163.33 N
a =F/m =163.33 /3.9 = 41.88 m/s^2
Net force makes an angle β with x-axis,
tan β = 93.74/133.74 = 0.7
The direction of the acceleration of the object is β = 35o above the x-axis

Answer 3

+y| / F
| /
| /
| /
/a_ _ _ _ _ +x
40N

This is pretty much what the picture looks like. Your answer was incorrect!

Answer 4

F1(x) = 40 N, F1(y) =0
F2 (x) =76•cosα = 53.74 N, F2 (y)= 76•sin α = 53.74 N ,
Net force F(x)=93.74 N, F(y) = 53.74 N.
F = sqrt(53.74^2 + 93.74^2) = 108 N
a =F/m =108 /3.9 = 27.7 m/s^2
Net force makes an angle β with x-axis,
tan β = 53.74/93.74 = 0.57
The direction of the acceleration of the object is β = 29.8o above the x-axis