Answer:
[tex]v=18.89\frac{m}{s}[/tex]
Explanation:
From the free-body diagram for the car, we have that the normal force has a vertical component and a horizontal component, and this component act as the centripetal force on the car:
[tex]\sum F_x:Nsin(20^\circ)=ma_c(1)\\\sum F_y:Ncos(20^\circ)=mg(2)[/tex]
Solving N from (2) and replacing in (1):
[tex]N=\frac{mg}{cos(20^\circ)}\\(\frac{mg}{cos(20^\circ)})sin(20^\circ)=ma_c\\gtan(20^\circ)=a_c[/tex]
The centripetal acceleration is given by:
[tex]a_c=\frac{v^2}{r}[/tex]
Replacing and solving for v:
[tex]gtan(20^\circ)=\frac{v^2}{r}\\v=\sqrt{grtan(20^\circ)}\\v=\sqrt{9.8\frac{m}{s^2}(100m)tan(20^\circ)}\\v=18.89\frac{m}{s}[/tex]
