- #1
ayans2495
- 58
- 2
- Homework Statement
- What is the mass of the object?
- Relevant Equations
- F = ma
Hiking the Appalachian Trail: A Beginner's Guide
- Thread starter ayans2495
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The discussion revolves around the concept of net force in relation to weight and mass calculations, specifically in the context of drag force. It highlights that a horizontal function does not always indicate zero net force, but in this case, it does due to the constant drag force. The conversation also references Newton's second law to clarify this relationship. Additionally, there is a mention of an unanswered question, suggesting a need for further engagement on the topic. Understanding the nuances of force dynamics is essential for accurate problem-solving in physics.
- #2
ayans2495
- 58
- 2
My answer: Horizontal function implies zero net force. Therefore, weight is 160 N. Thus, mass is 16 kg such that the acceleration due to gravity is 9.8 N/kg. Am I right?
- #3
Delta2
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Your answers are correct but your reasoning is not entirely correct or it is incomplete I would say. The part of the graph where it becomes horizontal doesn't necessarily imply that the net force is zero, however it is implying it in this specific case of drag force being constant. Why? What's the special thing about drag force and how this combined with Newton's second law implies that the net force is zero.
- #4
Delta2
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You haven't answer question 25 btw but ok its fairly easy after answering the first two questions.
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...
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