Respuesta :
Answer:
a)  W = - 6.825 J,  b) θ = 1.72 revolution
Explanation:
a) In this exercise the work of the friction force is negative and is equal to the variation of the kinetic energy of the particle
     W = ΔK
     W = K_f - K₀
     W = ½ m v_f² - ½ m v₀²
     W = ½ 0.325 (5.5² - 8.5²)
     W = - 6.825 J
b) find us the coefficient of friction
Let's use Newton's second law
      fr = μ N
y-axis (vertical) Â N-W = 0
      fr = μ W
work is defined by
       W = F d
the distance traveled in a revolution is
       d₀ = 2π r
       W = μ mg d₀ = -6.825
      μ = [tex]\frac{ -6.825}{d_o \ mg}[/tex]
       Â
The total work as the object stops the final velocity is zero v_f = 0
     W = 0 - ½ m v₀²
     W = - ½ 0.325 8.5²
     W = - 11.74 J
      μ mg d = -11.74
     Â
we subtitle the friction coefficient value
      ( [tex]\frac{-6.8525 }{d_o mg}[/tex]) m g d = -11.74
        6.825  [tex]\frac{d}{d_o}[/tex] = 11.74
        d = 11.74/6.825  d₀
        d = 1.7201  2π 0.400
        d = 4.32 m
this is the total distance traveled, the distance and the angle are related
       θ = d / r
       θ = 4.32 / 0.40
       θ = 10.808 rad
we reduce to revolutions
       θ = 10.808 rad (1rev / 2π rad)
       θ = 1.72 revolution
