Answer:
the minimum value of the coefficient of static friction between the ground and the cheetah's feet is 1.94
Explanation:
Given that ;
the top speed of Cheetahs is almost 60 mph
In cornering abilities ; the maximum centripetal acceleration of a cheetah was measured to be = 19 m/s^2
The objective of this question is to determine the what minimum value of the coefficient of static friction between the ground and the cheetah's feet is necessary to provide this acceleration?
From the knowledge of Newton's Law;
we knew that ;
Force F = mass m × acceleration a
Also;
The net force [tex]F_{net}[/tex] = frictional force [tex]\mu_k mg[/tex]
so we can say that;
m×a = [tex]\mu_k mg[/tex]
where;
the coefficient of static friction [tex]\mu_k[/tex] is:
[tex]\mu_k = \dfrac{m*a}{m*g}[/tex]
[tex]\mu_k = \dfrac{a}{g}[/tex]
[tex]\mu_k = \dfrac{19 \ m/s^2}{9.81 \ m/s^2}[/tex]
[tex]\mathbf{\mu_k}[/tex] = 1.94
Hence; the minimum value of the coefficient of static friction between the ground and the cheetah's feet is 1.94
