Question

A 2 kg block slides along a horizontal tabletop. A horizontal applied force of 12 N and a vertical applied force of 15 N act on the block. If the coefficient of kinetic friction between the block and the table is 0.2, what is the frictional force exerted on the block?

Answer 1

The block slides horizontally along the table, so the net vertical force is zero. By Newton's second law,

15 N + n + (-w) = 0

where n = magnitude of the normal force and w = weight of the block. We then find

n = w - 15 N

n = (2 kg) (9.8 m/s²) - 15 N

n = 4.6 N

The frictional force has a magnitude f that is proportional to n by a factor of µ = 0.2, such that

f = µ n

f = 0.2 (4.6 N)

f = 0.92 N

with a direction opposite the direction of the block's motion.