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.

your answer is, orbit around the nucleus

note: the given question is incomplete.

complete question is given below.

which is an example of kinetic energy?

a. the energy stored in ethanol  

b. a hockey puck sliding across the ice  

c. a compressed spring  

d. a ball sitting at the top of a ramp

solution: the kinetic energy is the energy posses by the virtue of the motion of the object.

in the given options, a, c and d are the example of potential energy because it is the energy posses by the virtue of the rest of the object. but option b has the kinetic energy as the object is in motion.

hence, a hockey puck sliding across the ice   is the apex example of kinetic energy.

so, the option (b) a hockey puck sliding across the ice   is the correct option.