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A 25 kg chair initially at rest on a horizontal floor requires a 165 N horizontal force to set it in motion. Once the chair is in motion, a 127 N horizontal force keeps it moving at a constant velocity.​

Find the coefficient of static friction between the chair and the floor.​

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b. Find the coefficient of kinetic friction between the chair and the floor.​

Sagot :

Lanuel

a. The coefficient of static friction between the chair and the floor is 0.67.

b. The coefficient of kinetic friction between the chair and the floor is 0.52.

Given the following data:

  • Mass of chair = 25 kg
  • Horizontal force A = 165 N
  • Horizontal force B = 127 N

Scientific data:

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

a. To find the coefficient of static friction between the chair and the floor:

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

[tex]F_s = uF_n = umg[/tex]

Where;

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

Making μ 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{165}{25 \times 9.8} \\\\u=\frac{165}{245}[/tex]

μ = 0.67

b. To find the coefficient of kinetic friction between the chair and the floor:

Note: The normal force would be the same as above.

[tex]u_k = \frac{F_k}{F_n} \\\\u_k = \frac{127}{245}\\\\u_k =0.52[/tex]

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

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