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.

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.

im srry but ur wrong the answer is d a smooth rock, some red dirt, and an empty bird's nest.