What force must be applied to push a carton weighing 350 N up a 19° incline, if the coefficient of kinetic friction is 0.45? Assume the force is applied parallel to the incline and the velocity is constant. (Enter the magnitude of the force.)

To find the force that must be applied to push the carton up the incline, we need to consider the forces acting on the carton.

First, let's break down the components of the weight of the carton. The weight of the carton can be divided into two components: the force acting parallel to the incline (mg*sinθ) and the force acting perpendicular to the incline (mg*cosθ), where m is the mass of the carton and θ is the angle of the incline.

Since we know the weight of the carton is 350 N, we can find the mass by dividing the weight by the acceleration due to gravity, g (approximately 9.8 m/s^2):
m = 350 N / 9.8 m/s^2
m ≈ 35.71 kg

Next, we can determine the force of kinetic friction acting against the motion of the carton. The force of kinetic friction (fk) can be found by multiplying the coefficient of kinetic friction (μk) by the normal force (N), where N is the force perpendicular to the incline:
fk = μk * N

The normal force (N) can be found by multiplying the weight of the carton by the cosine of the angle of the incline:
N = mg * cosθ

In this case, since the velocity is constant, the applied force must overcome the force of kinetic friction. Therefore, the force required to push the carton up the incline is equal to the force of kinetic friction:
Force applied = fk

Now, let's calculate the force required to push the carton up the incline:

First, calculate the normal force:
N = (35.71 kg * 9.8 m/s^2) * cos(19°)
N ≈ 330.2 N

Next, calculate the force of kinetic friction:
fk = 0.45 * 330.2 N
fk ≈ 148.59 N

Therefore, the magnitude of the force that must be applied to push the carton up the incline is approximately 148.59 Newtons.