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A 2000 kg roller coaster is at the top of a loop with a radius of 24 m. If its speed is 18 m/s at this point, what force does it exert on the track

Sagot :

boffeemadrid

Answer:

[tex]46620\ \text{N}[/tex]

Explanation:

m = Mass of roller coaster = 2000 kg

r = Radius of loop = 24 m

v = Velocity of roller coaster = 18 m/s

g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

Normal force at the point will be

[tex]N-mg=\dfrac{mv^2}{r}\\\Rightarrow N=\dfrac{mv^2}{r}+mg\\\Rightarrow N=\dfrac{2000\times 18^2}{24}+2000\times 9.81\\\Rightarrow N=46620\ \text{N}[/tex]

The force exerted on the track is [tex]46620\ \text{N}[/tex].

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