Answer:
[tex](a) v=16.2m/s\\(b) u_{k}=0.237[/tex]
Explanation:
Given data
[tex]r=100.0m\\\alpha =15.0^{o}\\v_{2}=20.0km/h=5.6m/s[/tex]
For Part (a) Speed
The speed v is given by
[tex]tan(\alpha )=\frac{v^2}{r.g}\\ v^2=tan(\alpha )*r.g\\v=\sqrt{tan(\alpha )*r.g}\\ v=\sqrt{tan(15)*100.0m*9.81m/s^2}\\ v=16.2m/s[/tex]
For Part (b) minimum coefficient of friction
To determine the friction of coefficient we know that friction force f is given by:
[tex]f=u_{k}.N=u_{k}.m.g[/tex]
The first centripetal force Fc₁ is given by:
[tex]F_{c1}=ma_{c1}\\F_{c1}=\frac{m.v_{1}^2}{r}\\ F_{c1}=\frac{m.(16.2m/s)^2}{100m}\\ F_{c1}=m.2.62[/tex]
The second centripetal force Fc₂ is given by:
[tex]F_{c2}=ma_{c2}\\F_{c2}=\frac{m.v_{2}^2}{r}\\ F_{c2}=\frac{m.(5.6m/s)^2}{100m}\\ F_{c2}=m.0.3[/tex]
The additional friction force is given by:
[tex]f=|F_{c1}-F_{c2}|\\m*u_{k}*g=|m*2.62-m*0.3|\\m*u_{k}*g=m|2.62-0.3|\\u_{k}*g=|2.62-0.3|\\u_{k}=\frac{2.62-0.3}{9.8m/s^2}\\u_{k}=0.237[/tex]
