Max speed of roller coaster at bottom of drop

Click For Summary
SUMMARY

The maximum speed of a roller coaster at the bottom of a drop from a height of 77.7 meters can be calculated using work and energy equations, specifically Wg = -(mgyf - mgyi) and Wnet = (1/2 mvf² - 1/2 mvi²). Assuming no friction and an initial velocity of 0 m/s at the top, the equations can be set equal to each other to solve for the final velocity (vf). The mass of the roller coaster cancels out, allowing for a straightforward calculation of speed based solely on gravitational potential energy and kinetic energy principles.

PREREQUISITES
  • Understanding of gravitational potential energy (mgh)
  • Familiarity with kinetic energy equations (1/2 mv²)
  • Basic knowledge of work-energy principles
  • Ability to manipulate algebraic equations
NEXT STEPS
  • Study the conservation of mechanical energy in physics
  • Learn about the effects of friction on roller coaster dynamics
  • Explore advanced topics in energy transformations
  • Investigate real-world applications of work-energy principles in engineering
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking for practical examples of energy conservation in real-world scenarios.

pb23me
Messages
204
Reaction score
0

Homework Statement


The top of the roller coaster is 77.7m above the earth. From this height what speed can be reached when it reaches the bottom of the drop? This problem must be solved using work and energy equations not kinematic equations..


Homework Equations


Wg=-(mgyf-mgyi)
Wnet=(1/2mv2f-1/2mv2i)


The Attempt at a Solution

I set the two equations equal to each other-assuming no friction-and tryed to solve for vf. However i can't see how to come up with that if i don't know the initial velocity?
 
Last edited:
Physics news on Phys.org
Normally roller coasters stop at the top of a steep fall, so I would assume that its initial velocity is 0 m/s. Otherwise you do not have enough information to get the initial velocity.
 
Ha thriller i remember these questions man, I'm like oh no how on Earth can i do this when i don't know the mass?! But setting E=05mv^2 and E=mgh equal to each other sound like a grand plan if you want to get rid of those cheeky masses! Cuz don't forget they should equal each other at the bottom of the slope in this imaginary magical world where the drop is perfectly vertical and there are no energy losses whatsoever. And yes rock.freak starting with u=0 is grand also. You've probs already solved it by now eh? Gosh darnit i wanted to sound smart. :( ;)
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
867
Replies
8
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 10 ·
Replies
10
Views
6K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 15 ·
Replies
15
Views
5K
  • · Replies 3 ·
Replies
3
Views
7K
  • · Replies 15 ·
Replies
15
Views
7K
Replies
12
Views
3K