contestada

A 65.0 kg skier slides down a
37.2° slope with uk = 0.107.
If the slope is 42.0 m long, and
the skier starts from rest, how
fast is she going at the bottom?
(Unit = m/s)

Respuesta :

She is going at a velocity of 20.63m/s to the bottom.

Friction is a resistive force between a body and the surface it is moving over in mechanics. It is computed mathematically using the friction coefficient and normal force product.

The mass is:  m = 65kg

The coefficient of friction is:  μk = 0.107

The angle is:  θ = 37.2°

W = mg

W₁ = W sin θ = mg sin θ

W = W cos θ = mg cos θ

[tex]F_{N} = W = mgcos[/tex]θ

[tex]F_{f} =[/tex] µ[tex]F_{N}[/tex] = µmg cosθ

[tex]F_{net} = W_{1} - F_{f}[/tex]

       [tex]= mg sin[/tex]θ - µmg cosθ

[tex]F_{net} = ma[/tex]

[tex]ma =[/tex] mg sinθ - µmg cosθ

a = g sinθ - µmg cosθ

  = 9.8 sin (37.2) - (0.107) 9.8 cos(37.2)

  = 9.8 × 0.605 - 0.107 × 9.8 × 0.796

  = 5.929 - 0.834

  = 5.095 [tex]m^2/s[/tex]

∴ Δd = [tex]\frac{1}{2} a ([/tex][tex]t^2[/tex][tex])[/tex]

  [tex]t^2[/tex] = 2Δd / a

  [tex]t^2=[/tex] 2 × 42.0 / 5.095

  [tex]t^2 =[/tex] 16.48

  [tex]t=4.05[/tex]

∴ [tex]v_{2} = v_{1} + at[/tex]

      [tex]= 0+ 5.095 (4.05)[/tex]

      [tex]= 20.63 m/s[/tex]

Therefore, she is going at a velocity of  20.63m/s to the bottom.

Learn more about friction here:

https://brainly.com/question/23161460

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