Physics Help Please Thank you?

You drag a suitcase of mass 7.2 kg with a force of F at an angle 46.1

◦with respect to the horizontal along a surface with kinetic coefficient of friction 0.32.

The acceleration of gravity is 9.8 m/s

If the suitcase is moving with constant velocity 1.83 m/s, what is F?

Answer in units of N.

I thought the way to approach this was to set F*x=(1/2)*m*v^2

NO distance is given!

Because the suit case is on an inclined surface, the weight of the suitcase has 2 components, Force parallel and Force perpendicular to the surface.

1) Force parallel to the inclined surface = mass * g * sin θ = 7.2 * 9.8 * sin 46.1°

F parallel = 50.8 N

2) Force perpendicular to the inclined surface = mass * g * cos θ

Force of friction = µ * Force perpendicular

Force of friction = µ * mass * g * cos θ = 0.32 * 7.2 * 9.8 * cos 46.1°

Force of friction = 15.7 N

I assume that you are dragging the suitcase up the incline.

Force parallel is DOWN the incline.

The force of friction is always in the direction opposite to the direction of the velocity, so the direction of the force of friction is DOWN the incline.

The sum of the 2 forces = 50.8 + 15.7 = 66.5 N

The sum of the 2 forces is causing the suitcase to DECELERATE, as you pull the suitcase UP the incline.

The 3rd force = F, is the only force causing the suitcase to ACCELERATE.

Total force = F – 66.5 = mass * acceleration

Since the suitcase is moving with CONSTANT velocity of 1.83 m/s, the acceleration of the suitcase = 0 m/s^2!!

F – 66.5 = 0

F = 66.5 N

I assume you and the suitcase were moving at 1.83 m/s, before this problem even started!!

Sneaky problem!