Friction of a sliding mass on a slope

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The discussion centers on understanding the friction of a sliding mass on a slope, emphasizing the role of kinetic friction in determining the mass's movement. Participants highlight the importance of calculating the normal force, expressed as N=mg*cos(theta), which is crucial for analyzing static friction. Questions arise about how kinetic friction relates to the normal force and the forces acting on the mass as it slides down the ramp. The necessity of applying free body diagrams is mentioned to visualize the forces at play. Overall, the conversation stresses the need for a solid grasp of these concepts to solve the problem effectively.
raspberrypienjoyer
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Homework Statement
Under which condition will the sliding stop? Show how you found this condition.
When the sliding stops, what distance will be traveled since t=0?
Relevant Equations
At t=0, its velocity is V0
hw5.png

Could you please help me with this? I guess move of the mass will be determined with kinetic friction.
 
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What are your thoughts? How does this relate to what you have been studying?

You must make a serious attempt at the problem yourself before we can help.
 
PeroK said:
What are your thoughts? How does this relate to what you have been studying?

You must make a serious attempt at the problem yourself before we can help.
Static friction is N=mg*costheta as far as I know.
 
raspberrypienjoyer said:
Static friction is N=mg*costheta as far as I know.
That's the normal force due to gravity. How is kinetic friction related to that.

Also, have you learned about free body diagrams?
 
What force is sliding it down the ramp and what is that force equal to if the block is stopped?
 
Thread 'Chain falling out of a horizontal tube onto a table'

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...
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