Find the net force of the box and the acceleration. 10 points. Will give brainliest.

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

a displacement distance is size of displ

amplitude

hope this