Can a sudden blow truly change potential energy?

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A sudden blow does not change potential energy because potential energy is dependent on position, which does not change instantaneously. The duration of the blow is negligible, meaning there is insufficient time for any significant displacement to occur. As a result, while the speed may increase, the position relative to the pivot remains effectively unchanged. Therefore, the potential energy remains constant during such a brief interaction. This understanding clarifies why potential energy is unaffected by a sudden force.
Lil123
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Homework Statement
A 2-kg mass on a frictionless table is attached to one end of a
massless spring. The other end of the spring is held by a frictionless
pivot. The spring produces a force of magnitude 3r newtons
on the mass, where r is the distance in meters from the pivot to the
mass. The mass moves in a circle and has a total energy of 12 J.The mass is struck by a sudden sharp blow, giving it
instantaneous velocity of 1 m/s radially outward.
Relevant Equations
E =kinetic energy +potential energy
Why there will be no change in potential energy due to this sharp blow ? Potential energy is a function of position , so won't this position get changed due to this sharp blow?Please help!
 

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Lil123 said:
Why there will be no change in potential energy due to this sharp blow ? Potential energy is a function of position , so won't this position get changed due to this sharp blow?
What is your reasoning for saying that "there will be no change in potential energy"?
 
There is no instantaneous change in its PE, since its distance from the pivot does does not change instantly; only its speed from the pivot does.
 
@haruspex So, is it correct to say that mass doesn't have enough time to change its potential energy due to this sudden blow
 
Lil123 said:
@haruspex So, is it correct to say that mass doesn't have enough time to change its potential energy due to this sudden blow
Yes. The intent of "sudden" is that the duration is negligible. If it lasts ##\Delta t## and produces a velocity ##v## then the displacement during the blow is, roughly, ##v\Delta t##. If ##\Delta t## is negligible then so is the displacement.
 
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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