Max speed of roller coaster at bottom of drop

AI Thread Summary
To determine the maximum speed of a roller coaster at the bottom of a drop from a height of 77.7 meters, the work-energy principle is applied, equating gravitational potential energy to kinetic energy. Assuming no friction and an initial velocity of 0 m/s at the top, the equations Wg = -(mgyf - mgyi) and Wnet = (1/2 mvf^2 - 1/2 mvi^2) are utilized. By eliminating mass from the equations, the problem simplifies, allowing for the calculation of final velocity at the bottom of the drop. The discussion highlights the importance of energy conservation in solving the problem without needing the mass of the roller coaster. The approach emphasizes the theoretical scenario of a frictionless environment for accurate results.
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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?
 
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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. :( ;)
 
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