Qusetion of kinetic and static friction force

A box of mass m = 2.5 kg is released from rest near the top of the incline. The angle of the incline is 35 degrees above the horizontal. The coefficients of static and kinetic friction between the box and the incline are u_s = 0.20 and u_k = 0.15, respectively. What will be the magnitude of the frictional force on the box once it is released?

Explanation:

In order to find the frictional force, you must first decide whether the block will be able to slide down (check if the normal force multiplied by the co-efficient of static friction is less than the force of gravity acting parallel to the incline (mgsin(theta)). When I did it, the parallel force of gravity was around 14N, while the force of static friction was 4N. Now, we can establish that the block slides down the hill. All we have to do now is multiply the normal force by the co-efficient of kinetic friction, which gives us approximately 3N

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explanation:

1/0.14nm = 10^10/1.4 a lot