Respuesta :
Answer:
For no friction condition there is no range of speed only on possible speed is 41.1 km/h
while for the case of 0.300 friction coefficient the range of speed is from 6.5 m/s to 15.75 m/s
Explanation:
When there is no friction on the turn of road then the centripetal force is due to the component of Normal force only
So we will have
[tex]Ncos\theta = mg[/tex]
[tex]Nsin\theta = \frac{mv^2}{R}[/tex]
so we have
[tex]tan\theta = \frac{v^2}{Rg}[/tex]
[tex]\theta = tan^{-1}\frac{v^2}{Rg}[/tex]
v = 41.1 km/h = 11.42 m/s
[tex]\theta = tan^{-1}\frac{11.42^2}{28(9.8)}[/tex]
[tex]\theta = 25.42^o[/tex]
so here in this case there is no possibility of range of speed and only one safe speed is possible to take turn
Now in next case if the coefficient of static friction is 0.300
then in this case we have
[tex]v_{max} = \sqrt{(\frac{\mu + tan\theta}{1 - \mu tan\theta})Rg}[/tex]
[tex]v_{max} = \sqrt{(\frac{0.3 + 0.475}{1 - (0.3)(0.475)})(28 \times 9.8)}[/tex]
[tex]v_{max} = 15.75 m/s[/tex]
Similarly for minimum speed we have
[tex]v_{min} = \sqrt{(\frac{\mu - tan\theta}{1 + \mu tan\theta})Rg}[/tex]
[tex]v_{min} = \sqrt{(\frac{-0.3 + 0.475}{1 + (0.3)(0.475)})(28 \times 9.8)}[/tex]
[tex]v_{min} = 6.5 m/s[/tex]
So the range of the speed is from 6.5 m/s to 15.75 m/s
