Answer :
Answer:
14.38 m/s
Explanation:
We are given that
The coefficient of static friction between the road and the tires on a car=[tex]\mu_s=0.586[/tex]
Radius of curve=r=36 m
We have to find the speed will put the car on the verge of sliding as it rounds a level curve.
Magnitude of acceleration of car=[tex]\frac{v^2}{r}[/tex]
By Newton's second law
[tex]f_s=\frac{mv^2}{r}[/tex]
Where [tex]f_s=\mu_s mg[/tex]=Friction force
If car does not slip then
[tex]\frac{v^2}{r}\leq \mu_sg[/tex]
[tex]v^2\leq \mu_srg[/tex]
The maximum speed with which the car can round the curve without slipping
[tex]v^2_{max}=\mu rg[/tex]
[tex]v_{max}=\sqrt{\mu_srg}[/tex]
Substitute the values and taking g=[tex]9.8m/s^2[/tex]
[tex]v_{max}=\sqrt{0.586\times 36\times 9.8}=14.38m/s[/tex]
Hence, the speed of the car on the verge of sliding as it round a level curve=14.38 m/s