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the speed of the rollercoaster is 22.1 m/s at the position marked r1. the radius at this point is 13.9 m. calculate the size of the force required by the track for a coaster of mass 2,498 kg to negotiate this turn. report this value as a whole number.

User Aeoliyan
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Final answer:

The force required by the track for the coaster to negotiate this turn is approximately 9761 N when the speed of the rollercoaster is 22.1 m/s at the position marked r1. the radius at this point is 13.9 m. calculate the size of the force required by the track for a coaster of mass 2,498 kg to negotiate this turn.

Step-by-step explanation:

To calculate the force required by the track for the rollercoaster to negotiate the turn, we need to use the centripetal force formula, which is given by:

Fc = (m*v^2) / r

where Fc is the centripetal force, m is the mass of the rollercoaster, v is its speed, and r is the radius of curvature.

Given that the speed of the rollercoaster is 22.1 m/s and the radius of curvature is 13.9 m, we can substitute these values into the formula:

Fc = (2498 kg * (22.1 m/s)^2) / 13.9 m

= 9761 N

Therefore, the force required by the track for the coaster to negotiate this turn is approximately 9761 N.

User MikeTheReader
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