A 792 kg rollercoaster car is on theinside of a loop of radius 18.7 m.At the top of the loop, the car movesat 17.3 m/s. What is the normalforce on the car at that point?(Unit = N)Enter

mass = 792 kg
r = 18.7 m
v = 17.3 m/s
Centripetal force is a force in a circle with a center. Generally, the normal force plus the weight of the car(mg) equals the net force(centripetal force). Therefore,
[tex]\begin{gathered} f=\frac{mv^2}{r^{}} \\ f=\frac{792\times17.3}{18.7} \\ f=\frac{237037.68}{18.7} \\ f_c=12675.8117647N \end{gathered}[/tex]To find the normal force
[tex]\begin{gathered} mg+\text{normal force=732.71} \\ 792\times9.8+normal\text{ force=732}.71 \\ 7761.6+normal\text{ force=}12675.8117647 \\ \text{normal force=}4914.21176471N \end{gathered}[/tex]