How Much Should a Spring Be Compressed for a Roller Coaster to Complete a Loop?

AI Thread Summary
To determine how much a spring must be compressed to launch an 840 kg roller coaster car through a loop with a radius of 6.2 m, the conservation of energy principle is applied. The initial potential energy from the spring must equal the total energy at the top of the loop, which includes gravitational potential energy and kinetic energy. The calculations suggest that the height at the top of the loop is 3.1 m, leading to the conclusion that the spring's compression must provide sufficient energy to achieve this height. The discussion emphasizes the importance of equating initial and final energy states to solve for the required spring compression. Understanding these energy transformations is crucial for ensuring the roller coaster remains on track.
stella77
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Conservation of Energy please help!

Homework Statement



An 840 kg roller coaster car is launched from a giant spring of constant k=31kN/m into a frictionless loop-the-loop track of radius 6.2m, what is the minimum amount that the spring must be compressed if the car is to stay on the track?

Homework Equations



v=n+mg = (mv^2)/r
v= √(gr)

mgh=0.5mv^2

The Attempt at a Solution



mgh=0.5mv^2
mgh=0.5m √(gr)^2
mgh=0.5mgr
h = 0.5r
h= 3.1 m

and I am stuck after this... please help...

 
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Your velocity is correct.

It has initial energy which should be equal to the energy at the top point of the loop (what energies are present here?).
Find it's initial energy and set that equal to its final energy.
 


the initial energy is the potential energy right??
 


Yes and then there are two energies present at the top of the loop.
 

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