Answer:
0.04
Explanation:
From
[tex]ma=\mu mg[/tex]
Here m is mass, a is acceleration, [tex]\mu[/tex] is the coefficient of kinetic friction and g is acceleration due to gravity.
Making [tex]\mu[/tex] the subject of the formula
[tex]\mu=\frac {a}{g}[/tex]
From kinematics, we know that
[tex]v^{2}-u^{2}=2as[/tex]
Here, v is final speed, u is initial speed, a is acceleration and s is distance moved.
Making a the subject of the formula then
[tex]a=\frac {v^{2}-u^{2}}{2s}[/tex]
Since it stops, the final velocity is 0 while the initial speed is given as 13.7 m/s
Substituting 0 for v, 13.7 for u, 216.9 m for s then
[tex]a=\frac {0^{2}-13.7^{2}}{2\times 216.9}=-0.432664822 m/s^{2}[/tex]
Taking g as [tex]9.81 m/s^{2}[/tex]then using the formula
[tex]\mu=\frac {0.432664822}{9.81}=0.044104467\approx 0.04[/tex]
