Answer:
21.7 m/s
Explanation:
To solve this problem, we have to write the equations of motion along the two directions: horizontal and vertical.
Horizontal direction:
[tex]N sin \theta + \mu N cos \theta = m\frac{v^2}{r}[/tex]
Vertical direction:
[tex]N cos \theta - \mu N sin \theta - mg = 0[/tex]
where:
N is the normal reaction on the car
[tex]\theta=20^{\circ}[/tex] is the banking angle of the road
[tex]\mu=0.58[/tex] is the coefficient of friction of concrete
m is the mass of the car
v is the speed of the car
r = 40 m is the radius of the curve
[tex]g=9.8 m/s^2[/tex]is the acceleration due to gravity
Combining the two equations together, we can find the maximum speed allowed for the car:
[tex]v=\sqrt{\frac{rg(sin \theta + \mu cos \theta)}{cos \theta - \mu sin \theta}}[/tex]
And substituting the data we have, we find:
[tex]v=\sqrt{\frac{(40)(9.8)(sin 20^{\circ} + (0.58) cos 20^{\circ} )}{cos 20^{\circ} - (0.58) sin 20^{\circ} }}=21.7 m/s[/tex]
