friction/force

A 50kg box is placed on an inclined plane making an angle of 30 degrees with the horizontal, If the coefficient of kinetic friction is 0.30, what is the resultant force on the box?

The forces will have two components: vertical and horizontal.
If we use a referential adjacent to the surface, then we'll have the x component of weight and friction force in the x axis and the normal force and y component of force, which are equal in magnitude, cancelling each other out.
If you write this using the second law of Newton you get:

Fx = m*a ⇔
Fy = 0 ⇔

⇔ m*g*sin(30) - µN = m*a ⇔
⇔m*g*cos(30) = N ⇔

⇔ m*g*sin(30) - µ*m*g*cos(30) = m*a ⇔
⇔ m*g*cos(30) = N ⇔

⇔ g*[ sin(30) - µ*cos(30) ] = a
⇔ m*g*cos(30) = N

Now you can substitute and get the values you wish.You know that µ=0.3 and g=9.8 m/s².