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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?

Respuesta :

LammettHash

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.

onyebuchinnaji

The frictional force exerted on the block is 0.92 N.

The given parameters;

  • mass of the block, m = 2 kg
  • horizontal force,  = 12 N
  • vertical force, = 15 N
  • coefficient of kinetic friction, μ = 0.2

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;

[tex]\Sigma F_y = 0\\\\15 + n - W = 0\\\\15 + n - (mg) = 0\\\\15 + n - (2\times 9.8) = 0\\\\15 + n - 19.6=0\\\\n - 4.6 = 0\\\\n = 4.6\ N[/tex]

The frictional force exerted on the block;

[tex]F_k = \mu \times n= 0.2 \times 4.6 = 0.92 \ N[/tex]

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

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