Answer:
[tex]\frac{0.065}{r}[/tex]
Explanation:
The maximum velocity of an object moving in a curve beyond which it will slide off the curve is given by the relationship in equation (1);
[tex]v=\sqrt{\mu gr}....................(1)[/tex]
where [tex]\mu[/tex] is the coefficient of friction between the object and the surface of the curve, g is acceleration due to gravity and r is the radius of the curve.
Given;
v = 0.8m/s
g = [tex]9.81m/s^2[/tex]
r = ?
[tex]\mu=?[/tex]
In order to solve for [tex]\mu[/tex], we can simply make it the subject of formula from equation (1) as follows;
[tex]v^2=\mu gr\\hence\\\mu=\frac{v^2}{gr}.................(2)[/tex]
since we were not given the value of r, we can just substitute other known values, then solve and leave the answer in terms of r.
Therefore;
[tex]\mu=\frac{0.8^2}{9.81r}[/tex]
[tex]\mu=\frac{0.64}{9.81r}\\\\\mu=\frac{0.065}{r}[/tex]
