Determining speed and height on a roller coaster

AI Thread Summary
The discussion focuses on calculating the speed and height at various points on a roller coaster using conservation of energy principles. The initial speed at point A is given as 50.0 cm/s, and the user successfully calculated the speed at point B as 44.3 m/s. However, discrepancies arise when calculating speeds at points C and E, with the user's answers significantly lower than those in the answer key. The conversation emphasizes the importance of correctly applying the conservation of energy equation, PE1 + KE1 = PE2 + KE2, and clarifies the user's misunderstanding regarding potential and kinetic energy at different points. Overall, the discussion highlights the need for careful application of energy conservation principles in solving roller coaster physics problems.
crushedcorn
Messages
7
Reaction score
0

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!
 
Physics news on Phys.org
You quoted the equation: PE1 + KE1 = PE2 + KE2
You used the equation: PE = KE
 
  • Like
Likes crushedcorn
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
 
Last edited:
  • Like
Likes crushedcorn
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
 
  • Like
Likes crushedcorn
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
 
Last edited:
  • Like
Likes crushedcorn
Back
Top