How Does Energy Conservation Apply to MCAT Physics Problems?

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Energy conservation principles are crucial for solving MCAT physics problems, as they help in understanding the relationships between kinetic and potential energy. The discussion highlights a specific question where the correct answer was identified as C, while another participant initially chose B. This indicates the importance of reviewing energy conservation concepts to avoid common mistakes. Participants emphasize the need for practice and comprehension of these principles to improve performance on the MCAT. Mastering energy conservation can significantly enhance problem-solving skills in physics-related questions on the exam.
Nick tringali
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Homework Statement
Question number 14. I figured it would have to be B because Te1=Te2. I feel like both b and c were correct. Why is it not B?
Relevant Equations
Pe=mgh
Ke=.5mv^2
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The book states the correct answer is C I chose B
 
I just figured this out by myself, thank you anyways!
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...