Answer:
Explanation:
Given
Radius of Track [tex]r_1=75 m[/tex]
coefficient of Static Friction [tex]\mu _s=0.78[/tex]
Here centripetal Force is Balanced by Friction Force
thus
[tex]\frac{mv^2}{r}=\mu _sg[/tex]
[tex]\frac{v^2}{r}=\mu _sg[/tex]
[tex]v=\sqrt{\mu _srg}[/tex]
[tex]v=\sqrt{0.78\times 75\times 9.8}[/tex]
[tex]v=23.94 m/s[/tex]
(b)For [tex]r_2=25 m[/tex]
[tex]\mu _s=0.12[/tex]
[tex]v=\sqrt{\mu _sr_2g}[/tex]
[tex]v=\sqrt{0.12\times 25\times 9.8}[/tex]
[tex]v=5.42 m/s[/tex]
