Answer:
4.18
Explanation:
Givens
The car's initial velocity [tex]v_{i}[/tex]= 0 and covering a distance Δx = 1/4 mi = 402.336 m in a time interval t = 4.43 s.
Knowns
We know that the maximum static friction force is given by:
[tex]f_{s_max} =[/tex]μ_s*n (1)
Where μ_s is the coefficient of static friction and n is the normal force.
Calculations
(a) First, we calculate the acceleration needed to achieve this goal by substituting the given values into a proper kinematic equation as follows:
Δx=[tex]v_{i} +\frac{1}{2} at^2[/tex]
a=41 m/s
This is the acceleration provided by the engine. Applying Newton's second law on the car, so in equilibrium, when the car is about to move, we find that:
[tex]f_{y}=n-mg=0\\ n=mg\\f_{x}=F-f_{s,max} =0\\ f_{s,max}=F=ma\\[/tex]
Substituting (3) into (1), we get:
[tex]f_{s,max}=[/tex] μ_s*m*g
Equating this equation with (4), we get:
ma= μ_s*m*g
μ_s=a/g
=4.18
