Answer:
The coefficient of static friction between the car and the track
u=0.572
Explanation:
We don't know the mass of the car or any other information so the acceleration is the reason to solve the friction coefficient
∑[tex]F=F_{f}=m*a_{t}[/tex]
As we know
[tex]F_{f}=u*F_{N}=u*m*g[/tex]
Also the center ward direction forces
[tex]F_{fc}=m*a_{c}[/tex]
[tex]a_{c}=\frac{v_{t}^2}{r}[/tex]
[tex]F_{fc}=m*\frac{v_{t}^2}{r}[/tex]
But now vt relation with the tangential acceleration
[tex]v_{t}=2*a_{t}*\frac{\pi }{r}[/tex]
replacing
[tex]F_{fc}=m*a_{t}*\frac{2\pi*r}{2r}[/tex]
[tex]F_{fc}=m*a_{t}*\pi[/tex]
So magnitude of the force can get by
[tex]F_{f}=\sqrt{(m*a_{t}*\pi)^{2}+(m*a_{t})^{2}}[/tex]
Get the factor to simplify
[tex]F_{f}=a_{t}*m*\sqrt{(1+\pi^2)}[/tex]
[tex]u*m*g=m*a_{t}*(\sqrt{1+\pi^2})[/tex]
Solve to u'
[tex]u=\frac{a_{t}}{g}*(\sqrt{1+\pi^2})[/tex]
[tex]u=\frac{17\frac{m}{s^2} }{9.8\frac{m}{s^2}}*(\sqrt{1+\pi^2})=0.572[/tex]
