The answer is A I believe. Hope this helps!
The formula
is used when the statement of second law of motion is applied to the problem. If the statement is applied to the problem, The fomula willbe used. If not, use the formula 
Answer:
38.6 N
2.57 m/s²
Explanation:
Draw a free body diagram of the box. There are four forces:
Weight force mg pulling down,
Normal force N pushing up,
Friction force Nμ pushing left,
and applied force P pulling at an angle 40°.
Sum of forces in the y direction:
∑F = ma
N + P sin 40° − mg = 0
N = mg − P sin 40°
The net force in the x direction is:
∑F = P cos 40° − Nμ
∑F = P cos 40° − (mg − P sin 40°) μ
∑F = P cos 40° − mgμ + Pμ sin 40°
∑F = P (cos 40° + μ sin 40°) − mgμ
Plugging in values:
∑F = (80 N) (cos 40° + 0.23 sin 40°) − (15 kg) (10 m/s²) (0.23)
∑F = 38.6 N
Net force equals mass times acceleration:
∑F = ma
38.6 N = (15 kg) a
a = 2.57 m/s²