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2. A roller coaster at an amusement park has a dip that bottoms out in a vertical circle of radius r. A passenger feels the seat of the car pushing upward on her with a force equal to five times her weight as she goes through the dip. Of r = 21.5 m, how fast is the roller coaster traveling at the bottom of the dip?

Respuesta :

Answer:[tex]v=\sqrt{4gr}[/tex]

Explanation:

Given

Passenger feels the seat of the car Pushing upward with a force equal to five times her weight

Therefore

[tex]N-mg=\frac{mv^2}{r}[/tex]

it is given [tex]N=5mg[/tex]

where N=normal reaction

[tex]5mg-mg=\frac{mv^2}{r}[/tex]

thus [tex]v=\sqrt{4gr}[/tex]

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