A2,493-kg car is moving down a road with a slope (grade of 14% at a constant speed of 13 m/s. what is the direction and magnitude of the frictional force? (define positive x in the forward direction, i. e., down the slope?

To calculate for the frictional force, we calculate first for the normal force (forcer perpendicular to the surface)
                               Fn = (2,493 kg) (9.8 m/s²) = 24243.51 N
Then, we multiply this value with the frictional constant which is not given in this problem. If the frictional coefficient is equal to x then, the frictional force is equal to 24243.51x. 

A) acceleration. this seems to be the one that makes sense.