Answer:
0.40
Explanation:
Given that :
the mass of the caer = 250 kg
initial speed = 14 m/s
final speed = 0 m/s
distance s = 25 m
Using the equation of motion
[tex]v^2 = u^2 + 2as[/tex]
making a the subject of the formula ; we have:
[tex]v^2-u^2 = 2as[/tex]
[tex]a= \dfrac{v^2-u^2 }{2 \ s}[/tex]
[tex]a= \dfrac{(0)^2-(14)^2 }{2 \ (25)}[/tex]
[tex]a= \dfrac{0-196 }{50}[/tex]
[tex]a= \dfrac{-196 }{50}[/tex]
a = -3.92 m/s²
However; the relation for the coefficient of the kinetic static friction can be expressed as:
[tex]f= \mu_k *mg= ma[/tex]
[tex]f= \mu_k *g= a[/tex]
[tex]f= \mu_k = \dfrac{a}{g}[/tex]
[tex]f= \mu_k = \dfrac{3.92}{9.8}[/tex]
[tex]f= \mu_k = 0.40[/tex]
