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.)

Wc = 350 N. = Wt. of carton.

Fc = 350N @ 19o. = Force of carton.
Fp = 350*sin19 = 113.95 N. = Force parallel to incline.
Fv = 350*cos19 = 330.9 N. = Force perpendicular to incline.

Fk = u*Fv = 0.45*330.9 = 148.9 N. = Force of kinetic friction.

Fap-Fp-Fk = m*a.
Fap-113.95-148.9 = m*0 = 0
Fap-262.85 = 0
Fap = 262.85 N. = Force applied.

To find the force that must be applied to push the carton up the incline, we need to consider the following factors:

1. Gravitational force: The weight of the carton is given as 350 N, which is the force pulling it downward due to gravity. This force acts vertically downwards.

2. Normal force: The normal force is the perpendicular force exerted by the incline on the carton. It counteracts the gravitational force and prevents the carton from falling through the incline. The normal force is equal in magnitude but opposite in direction to the gravitational force, so it is 350 N acting vertically upwards.

3. Friction force: The coefficient of kinetic friction is given as 0.45. This coefficient is multiplied by the normal force to calculate the magnitude of the friction force. The equation for the friction force is given by F_friction = μ_k * N, where μ_k is the coefficient of kinetic friction and N is the normal force.

4. Force parallel to the incline: This is the force that needs to be applied to overcome both the gravitational force and the friction force. This force acts parallel to the incline and is directed upward.

Now, let's calculate the force that needs to be applied:

Friction force = coefficient of kinetic friction * normal force
Friction force = 0.45 * 350 N
Friction force = 157.5 N

The force required to push the carton up the incline is equal to the sum of the gravitational force and the friction force:

Force parallel to incline = gravitational force + friction force
Force parallel to incline = 350 N + 157.5 N
Force parallel to incline = 507.5 N

Therefore, the magnitude of the force that must be applied to push the carton up the 19° incline is 507.5 N.