Answer:
mu = 0.56
Explanation:
The friction force is calculated by taking into account the deceleration of the car in 25m. This can be calculated by using the following formula:
[tex]v^2=v_0^2+2ax\\[/tex]
v: final speed = 0m/s (the car stops)
v_o: initial speed in the interval of interest = 60km/h
= 60(1000m)/(3600s) = 16.66m/s
x: distance = 25m
BY doing a the subject of the formula and replace the values of v, v_o and x you obtain:
[tex]a=\frac{v^2-v_o^2}{2x}=\frac{0m^2/s^2-(16.66m/s)^2}{2(25m)}=-5.55\frac{m}{s^2}[/tex]
with this value of a you calculate the friction force that makes this deceleration over the car. By using the Newton second's Law you obtain:
[tex]F_f=ma=(1490kg)(5.55m/s^2)=8271.15N[/tex]
Furthermore, you use the relation between the friction force and the friction coefficient:
[tex]F_f= \mu N=\mu mg\\\\\mu=\frac{F_f}{mg}=\frac{8271.15N}{(1490kg)(9.8m/s^2)}=0.56[/tex]
hence, the friction coefficient is 0.56
