ap physics question

Question

A block of mass 2 kg slides along a horizontal tabletop. A horizontal applied force of 12 N and a vertical

applied force of 15 N act on the block, as shown above. If the coefficient of kinetic friction between the

block and the table is 0.2, the frictional force exerted on the block is most nearly...

I found fg= 20, and thought fn was 20 too but was confused on the vertical applied force...

Answer 1

Draw the free body diagram

The applied horizontal force (F_A-H) = 12 N

The weight (W) of the mass id (mg = (2)(9.8) = 19.6 N

The applied vertical force (F_A-V) = 15 N

The frictional force (F_F) is determined by F_F = μF_N

Assume g = -9.8 m/s²

To calculate FN, utilize the fact that there is no acceleration happening in the y-direction therefore the sum of the forces in the y-direction equals zero:

∑F_y = 0

F_N - F_A-V - W = 0

F_N = F_A-V + W = 15 + 19.6 = 34.6 N

F_F = μF_N = (0.2)(34.6) = 6.9 N

If you assume g = -10 m/s², then W = 20 N and F_N = 35 N making F_F = 7 N