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a 3.98 kg block slides down a ramp with an angle of inclination of \\(4…

Question

a 3.98 kg block slides down a ramp with an angle of inclination of \\(42.1^{\circ}\\). if the block slides down with an acceleration of \\(1.43\text{ m/s}^2\\), what is the coefficient of kinetic friction?

\\(\bigcirc\\) 0.707
\\(\bigcirc\\) 0.457
\\(\bigcirc\\) 0.891
\\(\bigcirc\\) 0.610

Explanation:

Identify given values and target unknown

We are given:

  • Mass of the block, \(m = 3.98\text{ kg}\)
  • Angle of inclination, \(\theta = 42.1^\circ\)
  • Acceleration down the ramp, \(a = 1.43\text{ m/s}^2\)
  • Acceleration due to gravity, \(g = 9.80\text{ m/s}^2\)

We need to find the coefficient of kinetic friction, \(\mu_k\).

Resolve forces perpendicular to the incline

Using the Vector Resolution and Normal Force Calculation knowledge points

$$ LATEXBLOCK0 $$

Set up the equation of motion down the incline

Using the Newtons Second Law knowledge point

$$ LATEXBLOCK1 $$

Solve for the coefficient of kinetic friction

We can simplify the equation by dividing all terms by \(m\):

$$ LATEXBLOCK2 $$

Calculate the numerical value of the coefficient

Substituting the values:

$$ LATEXBLOCK3 $$

Answer:

  • (A) 0.707 (Correct answer)
  • (B) 0.457
  • (C) 0.891
  • (D) 0.610