Final answer:
The frictional force exerted on the 2.0 kg block is equal to the horizontal force applied, which is 12 N, since the block does not move despite the static frictional force potentially being as high as μ_s × N.
Step-by-step explanation:
The student asks about the frictional force exerted on a 2.0 kg block when a horizontal force of 12 N to the right and a vertical force of 15 N downwards are applied. Given the coefficient of static friction between the block and the table is 0.50, and that the block does not move, the frictional force is equal to the horizontal force applied, which is 12 N.
The normal force on the block is not just its weight (due to gravity), but also includes the additional vertical force. The normal force (N) equals the weight (mg, where m is mass and g is acceleration due to gravity) plus the vertical force applied. Here, the normal force would be N = (2.0 kg × 9.8 m/s²) + 15 N. However, since the block does not move, the maximum static frictional force (f_s) that could be exerted would be f_s = μ_s × N, where μ_s is the coefficient of static friction. This value represents the maximum frictional force before the block would start moving, but since we know the block remains static and the applied horizontal force is 12 N, the static frictional force must be countering the applied force exactly, resulting in a frictional force of 12 N as well.