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Roller coaster loop cart velocity

  • Thread starter raffish
  • Start date
in an amusement park ride a cart of mass 300kg and carring four passangers each of mass 60kg is dropped from a vertical height of 120m alonga path that leads into a loop-the-loop machine of radius 30m. The cart then enters a straight stretch from A to C where friction brings it to rest after a distance of 40m. (see attachment)

a) find the velocity of the cart at A
answer: mgh = (1/2)mv²
540 x 10 x 120 = (1/2) x 540 x v²
Thus, v = 48.99 m/s

b) find the reaction force from the seat of the cart onto a passanger at B.
answer: well at the top the net force will be euqual to the weight and the normal force, but what reaction force are they talking about here. i mean there isnt any force pushing onto the seat is there?

c) what is the acceleration experienced by the cart from A to C (assumed constant)?
answer: ???
 

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answer: well at the top the net force will be euqual to the weight and the normal force, but what reaction force are they talking about here. i mean there isnt any force pushing onto the seat is there?
The "reaction force" is the normal force. Of course there's a force pushing onto the seat--person and seat push against each other: that's the "normal" force. (Unless the speed is just right so that no normal force is needed to make it around the curve.) You'll need to apply Newton's 2nd law to solve for this force.

c) what is the acceleration experienced by the cart from A to C (assumed constant)?
answer: ???
Use kinematics--you have the velocity and the distance, use them to find the acceleration. (Depending upon the kinematic equations at your disposal, this can be done in one step or two.)
 

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