Answer:
c. 0.816
Explanation:
Let the mass of car be 'm' and coefficient of static friction be 'μ'.
Given:
Speed of the car (v) = 40.0 m/s
Radius of the curve (R) = 200 m
As the car is making a circular turn, the force acting on it is centripetal force which is given as:
Centripetal force is, ![F_c=\frac{mv^2}{R}](https://tex.z-dn.net/?f=F_c%3D%5Cfrac%7Bmv%5E2%7D%7BR%7D)
The frictional force is given as:
Friction = Normal force × Coefficient of static friction
![f=\mu N](https://tex.z-dn.net/?f=f%3D%5Cmu%20N)
As there is no vertical motion, therefore,
. So,
![f=\mu mg](https://tex.z-dn.net/?f=f%3D%5Cmu%20mg)
Now, the centripetal force is provided by the frictional force. Therefore,
Frictional force = Centripetal force
![f=F_c\\\\\mu mg=\frac{mv^2}{R}\\\\\mu=\frac{v^2}{Rg}](https://tex.z-dn.net/?f=f%3DF_c%5C%5C%5C%5C%5Cmu%20mg%3D%5Cfrac%7Bmv%5E2%7D%7BR%7D%5C%5C%5C%5C%5Cmu%3D%5Cfrac%7Bv%5E2%7D%7BRg%7D)
Plug in the given values and solve for 'μ'. This gives,
![\mu = \frac{(40\ m/s)^2}{200\ m\times 9.8\ m/s^2}\\\\\mu=\frac{1600\ m^2/s^2}{1960\ m^2/s^2}\\\\\mu=0.816](https://tex.z-dn.net/?f=%5Cmu%20%3D%20%5Cfrac%7B%2840%5C%20m%2Fs%29%5E2%7D%7B200%5C%20m%5Ctimes%209.8%5C%20m%2Fs%5E2%7D%5C%5C%5C%5C%5Cmu%3D%5Cfrac%7B1600%5C%20m%5E2%2Fs%5E2%7D%7B1960%5C%20m%5E2%2Fs%5E2%7D%5C%5C%5C%5C%5Cmu%3D0.816)
Therefore, option (c) is correct.