- #1
butterflycandy
- 1
- 0
- Homework Statement
- nevermind
- Relevant Equations
- V=IR
Completely lost :(
Electromagnetism Help: Get Unstuck Now!
- Thread starter butterflycandy
- Start date
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- Tags
- Electromagnatism
Click For Summary
The discussion revolves around a request for help with an electromagnetism problem related to power consumption of an appliance when switching from a 120 V outlet in Canada to a 240 V outlet in the UK. Participants emphasize the importance of showing effort in problem-solving before receiving assistance. The original problem statement was edited by the original poster, leading to confusion and a lack of clarity in the discussion. The thread ultimately concludes without a resolution due to the missing information and the OP's decision to edit their post. The conversation highlights the necessity of adhering to homework guidelines for effective assistance.
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...