Respuesta :
Answer:
The coefficient of static friction is [tex]\mu_s = 0.6897[/tex]
Explanation:
From the question we are told that
The tangential acceleration is [tex]a = 2.05 \ m/s^2[/tex]
Generally the circumference of the circle is mathematically represented as
[tex]C = 2 \pi r \\[/tex]
and given that the car travels one quarter of the circular path before it skids off the track , then the distance traveled along the track is mathematically represented as
[tex]d = \frac{1}{4 } * C = \frac{2\pi r}{4} = \frac{\pi r}{2}[/tex]
Generally from kinematic equations
[tex]v ^2 = u^2 + 2 * a * d[/tex]
Here u is the initial velocity of the car which is 0 m/s
while v is the velocity of the car that keeps it from skiing off the track
So
[tex]v ^2 = 0 + 2 * a * \frac{\pi r}{2}[/tex]
=> [tex]v = \sqrt{a \pi r}[/tex]
Generally the centripetal force acting on the car is mathematically represented as
[tex]F_c = \frac{m v^2}{r}[/tex]
=> [tex]F_c = \frac{m (\sqrt{a \pi r } )^2}{r}[/tex]
=> [tex]F_c = m \pi a[/tex]
Generally the frictional force between the track and the tires of the car is mathematically represented as
[tex]F_f = \mu_s * m * g[/tex]
Generally the tangential force acting on the car is mathematically represented as
[tex]F_f = m * a[/tex]
Generally the resultant force acting on the car is mathematically \
[tex]F_r = \sqrt{F_t + F_c}[/tex]
=> [tex]F_r = \sqrt{ (ma)^2 + (\pi ma)^2 }[/tex]
=> [tex]F_r = \sqrt{ m^2a^2+ \pi^2 m^2a^2 }[/tex]
=> [tex]F_r = \sqrt{(ma)^2 (1 + \pi^2) }[/tex]
=> [tex]F_r = ma \sqrt{ 1 + \pi^2 }[/tex]
Generally the point before the car skids off the track, frictional force is equal to the resultant force and this is mathematically represented as
[tex]ma \sqrt{ 1 + \pi^2 } = \mu_s * m * g[/tex]
=> [tex]\mu_s = \frac{a}{g} * \sqrt{1 + \pi^2}[/tex]
=> [tex]\mu_s = \frac{2.05}{9.8} * \sqrt{1 + 3.142 ^2}[/tex]
=> [tex]\mu_s = 0.6897[/tex]