june72
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A 4.0 kg block is resting on a rough horizontal table. The coefficient of static friction us is 0.60. The static friction between the block and the table before any attempted motion is to start is N. (round to the nearest integer)

Respuesta :

temdan2001

Answer: 24 N

Explanation:

Given that the

Mass m = 4 kg

Coefficient of friction (μ) = 0.60

To calculate the static friction between the block and the table before any attempted motion, you will use the formula; F = (μ)N

Where

F = static friction

N = normal reaction = mg

g = acceleration due to gravity = 9.8 m/^2

Substitutes m, g and (μ) into the formula

F = 0.6 × 4 × 9.8

F = 23.52 N

Therefore, the static friction between the block and the table before any attempted motion is 24 N approximately.

Lanuel

The static friction between the block and the table before any attempted motion is to start equals 23.52 Newton.

Given the following data:

  • Mass of block = 4.0 kg
  • Coefficient of static friction = 0.60.

To determine the static friction between the block and the table before any attempted motion is to start:

Mathematically, the force of static friction is given by the formula;

[tex]F_s = uF_n[/tex]   ...equation 1.

Where;

  • [tex]F_s[/tex] represents the force of static friction.
  • μ represents the coefficient of friction.
  • [tex]F_n[/tex]  represents the normal force.

But, [tex]F_n = mg[/tex]

[tex]F_s = umg[/tex]   ...equation 2.

Substituting the given parameters into eqn. 2, we have;

[tex]F_s = 0.60 \times 4 \times 9.8\\\\F_s = 23.52\; Newton[/tex]

Force of static friction = 23.52 Newton

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