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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? (use g = 9.8 m/s2)

Respuesta :

Olajidey

Answer:

the force it exert on track is

46600N

Explanation:

The car acceleration

a = V²/R

= (18.0m/s)²/(24.0)

= 13.5m/s²

net force

F(net) = m * a

         =  2000kg * 13.5m/s²

         = 27000N

F(net) = F₀ - F₁

where

F₀ = upward force

F₁ = downward force

F₁ = m * g

   = (2000kg) * (9.8m/s²)

   = 19600N

upward force which is F₀

F₀ = F(net) + F₁

    = 27000N + 19600N

    = 46600N

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