Determining speed and height on a roller coaster

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Homework Help Overview

The discussion revolves around a roller coaster physics problem involving the conservation of energy principles to determine speeds at various points and the height at a specific point, given certain initial conditions and parameters.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the application of energy conservation equations, questioning the calculations for speeds at points B, C, and E. There is discussion about the correct interpretation of potential and kinetic energy at different points, particularly regarding the initial conditions at point A.

Discussion Status

Participants are actively engaging with the problem, providing insights into the equations used and questioning the assumptions made in the calculations. Some guidance has been offered regarding the conservation of energy and the importance of initial conditions, but there is no explicit consensus on the correct approach yet.

Contextual Notes

There are indications of confusion regarding the application of energy conservation principles, particularly in distinguishing between total energy and changes in energy. The problem also involves a missing value for the speed at point D, which adds complexity to the discussion.

crushedcorn
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Homework Statement


One car (m=80.0 kg) tracks through the roller coaster in the following diagram. As it passes point A, it has a speed of 50.0 cm/s.

a) Determine the speed at points B, C, and E
b) If the speed at point D is _______, determine it's height (ignore friction)

Ch 5 Roller Coaster.jpeg


Homework Equations


PE1 + KE1=PE2 + KE2
PE=mgy
KE=1/2mv^2

The Attempt at a Solution


a) I correctly got the velocity for point B as 44.3 m/s (1/2mv^2=mgy, 1/2(80 kg)v^2=(80kg)(9.80m/s^2)(100m). However, when I apply the same formula to points C and E my answer doesn't match that of the answer key.

Point C: (80 kg)(9.80 m/s^2)(38 m)=1/2(80 kg)v^2, v=27.3 m/s. Answer key answer: 34.9 m/s

Point E: (80 kg)(9.80 m/s^s)(20 m)=1/2(80 kg)v^2, v=19.8 m/s. Answer key answer: 39.6 m/s

b) 1/2(80 kg)v^2=(80 kg)(9.80 m/s^2)y. I can simplify this equation but I don't know how to solve this as I would have 2 variables that I need to solve for.

Thank you for your wonderful brains on this!
 
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You quoted the equation: PE1 + KE1 = PE2 + KE2
You used the equation: PE = KE
 
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Don't get confused with ΔPE=ΔKE
 
PE1 + KE1=PE2 + KE2 becomes PE=KE because, for example, at point C KE1=0 and PE2=0, so I end up using the PE1=KE2 equation.

How am I confusing ΔPE with ΔKE?
 
KE1≠0 PE2≠0

You haven't used the fact that it's moving at 0.5ms-1 at A

You aren't confusing ΔKE with ΔPE. From your work it appears you are confusing ΔKE=ΔPE with KE=PE
 
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Total energy conserved:

mgy1+1/2 mv12=mgy2+1/2 mv22

For point B:
1/2(80kg)v2+(80kg)g0=(80kg)(9.80ms-2)(100m)+1/2(80kg)(0.5ms-1)2

The only reason you got the first part "right" is because 1/2(80kg)(0.5ms-1) is really small compared to (80kg)(9.80ms-2)(100m) and so it didn't make a noticable difference to the answer
 
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You want to use:

PEA+KEA=(80kg)(9.80ms-2)(100m)+1/2(80kg)(0.5ms-1)2=PEx+KEx=mgy+1/2mv2

For all but the last parts, y is known

For the last part, v is known
 
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