Finding Acceleration for Box with Pulled Rope

[AsuraSky]

Homework Statement

A student decides to move a box of books into her dormitory room by pulling on a rope attached to the box. She pulls with a force of 80.0 N at an angle of 21.0° above the horizontal. The box has a mass of 25.0 kg, and the coefficient of kinetic friction between box and floor is 0.300. Find the acceleration of the box.

2. The attempt at a solution

Since the student is pulling with a 80.0 N force at a 21 degree angle, the x component would be

Fx = 80cos(21)

The frictional force would be .3(25)(9.8)

So therefore the net force in the x direction is

80cos(21) - .3(25)(9.8) = 1.18643412 N

To find acceleration divide the previous result by mass

1.18643412/25 = .047457 m/s²

I put this answer in; however the online homework marked me wrong. I've double checked my work many times and I don't see what I'm doing wrong. Any help would be appreciated

 

[rock.freak667]

Remember, the y-component of the force is Fsinθ.

So vertically, you will have

Fsinθ+N = mg where 'N' is your normal reaction. The presence of the y-component changes the value of the normal reaction 'N'.