Is the Roller Coaster Speed Calculation Correct?

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SUMMARY

The roller coaster speed calculation at the bottom of a dip with a radius of 25 meters requires a correct application of Newton's laws. The passenger experiences an upward normal force of 3 times her weight, leading to the equation 3Mg = mv²/r. After resolving the forces, the correct speed of the roller coaster at the bottom of the dip is calculated to be 27.12 m/s. The discussion emphasizes the importance of considering both the normal force and gravitational force to determine the net force acting on the passenger.

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Homework Statement



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 3.0 times her weight as she goes through the dip.If r=25m, how fast is the roller coaster traveling at the bottom of the dip?

The Attempt at a Solution



F=mv*2/r
3Mg=mv*2/r
3g=v*2/25
V=27.12ms-1
Is this the correct solution?
 
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No, you must look at the net force acting on the passenger before applying Newton's laws. You only looked at one of the forces. What is the other?
 
PhanthomJay said:
No, you must look at the net force acting on the passenger before applying Newton's laws. You only looked at one of the forces. What is the other?
But, isn't the magnitude of the force is equal to 3 times of the weight...where w is equal to mg
( base on the question)
 
the question does say that the force due to the seat is 3 times her weight. But she does have another force on her (as Phanthomjay said).
 
So, the net force is N-mg...where N(upward force) is equal to 3 times of mg(weight)...?
 
Thanks...
 

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