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?

The frictional force exerted on the block is 0.92 N.

The given parameters;

The frictional force exerted on the block is calculated by applying Newton's second law of motion, to determine the net vertical force on the block;

The frictional force exerted on the block;

Thus, the frictional force exerted on the block is 0.92 N.

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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.

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