Is the Roller Coaster Speed Calculation Correct?

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The discussion revolves around calculating the speed of a roller coaster at the bottom of a dip, where a passenger experiences a force of three times her weight. The initial calculation incorrectly considers only one force, leading to an erroneous conclusion. It is clarified that the net force acting on the passenger must be taken into account, which includes both the normal force and gravitational force. The correct approach involves recognizing that the upward normal force is three times the weight, requiring a reevaluation of the net force. Ultimately, understanding the net forces is crucial for accurately determining the roller coaster's speed.
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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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