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1. Suppose a 10 N force is applied to the side of a 4.0 kg block that is sitting on a table what would be the minimum value of the coefficient of static friction in order for the block to remain motionless?

Respuesta :

samuelonum1

Answer:

[tex]\mu= 0.25[/tex]

Explanation:

Given data

Force= 10N

mass= 4kg

r= 4*9.81

r= 39.24N

The expression for the force acting is expressed as

[tex]F=\mu*r[/tex]

substitute

[tex]\mu=F/r[/tex]

[tex]\mu= 10/39.24[/tex]

[tex]\mu= 0.25[/tex]

Lanuel

The minimum value of the coefficient of static friction in order for the block to remain motionless is 0.26.

Given the following data:

  • Force = 10 Newton
  • Mass of block = 4.0 kg

Scientific data:

  • Acceleration due to gravity = 9.8 [tex]m/s^2[/tex]

To determine the minimum value of the coefficient of static friction in order for the block to remain motionless:

Note: The force that is required to make the block motionless must be lesser than or equal to the force of static friction.

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

[tex]Fs = uFn = umg[/tex]

Where;

  • Fs represents the force of static friction.
  • μ represents the coefficient of friction.
  • [tex]F_n[/tex] represents the normal force.
  • g is the acceleration due to gravity.
  • m is the mass of an object.

Making u the subject of formula, we have:

[tex]u=\frac{F_s}{mg}[/tex]

Substituting the given parameters into the formula, we have;

[tex]u=\frac{10}{4.0 \times 9.8} \\\\u=\frac{10}{39.2}[/tex]

u = 0.26

Read more on force of static friction here: https://brainly.com/question/25686958

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