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M2 Hooke's law

Can someone describe me what's happening in part one when you release the particle?

(red writing is the mark scheme)

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Original post by Jian17
Can someone describe me what's happening in part one when you release the particle?

(red writing is the mark scheme)


The scenario is outlined below.

The red writing is considering conservation of energy approach. They treat the two halves of the string separarely (each having natural length 1.5m and modulus 45N). In fact the tension in the top string AP initially is zero. But tension does not matter in the energy approach.

The LHS of the problem takes the *stopping point* as the point with gravitational potential of zero (g.p.e = 0) which is exactly h metres below A. Hence, the initial GPE is the second term. The first term denotes the E.P.E. in the BP portion of the string.
The RHS simply says that there is only E.P.E coming from the string portion AP, since every other energy level is zero.



(edited 5 years ago)
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Jian17
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Original post by RDKGames
The scenario is outlined below.

The red writing is considering conservation of energy approach. They treat the two halves of the string separarely (each having natural length 1.5m and modulus 45N). In fact the tension in the top string AP initially is zero. But tension does not matter in the energy approach.

The LHS of the problem takes the *stopping point* as the point with gravitational potential of zero (g.p.e = 0) which is exactly h metres below A. Hence, the initial GPE is the second term. The first term denotes the E.P.E. in the BP portion of the string.
The RHS simply says that there is only E.P.E coming from the string portion AP, since every other energy level is zero.





Understood most of it but, why is the tension AP in the beginning 0? (So E.P.E =0) and afterwards there's only E.P.E coming from AP? Does it have to do with T in AP being much smaller than T in PB, therefore extension is only in the portion PB? thanks
Original post by Jian17
Understood most of it but, why is the tension AP in the beginning 0?


Because the distance AP = 1.5m which is exactly the natural length of the upper half of the string. So there is x=0 extension.

(So E.P.E =0) and afterwards there's only E.P.E coming from AP?


At any point when the particle moves down, the string AP is now extended hence it has E.P.E.

The question tells us to assume BP is slack when the particle comes to rest, so EPE in the lower portion is zero hence we ignore it.
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Jian17
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Original post by RDKGames
At any point when the particle moves down, the string AP is now extended hence it has E.P.E.

The question tells us to assume BP is slack when the particle comes to rest, so EPE in the lower portion is zero hence we ignore it.


Now I get it thanks, and in a scenario where theres E.P..E on both portions, would I add the individual E.P..E's for starting position and stopping position?
Original post by Jian17
Now I get it thanks, and in a scenario where theres E.P..E on both portions, would I add the individual E.P..E's for starting position and stopping position?


If a particle stops at a place >1.5m above B then you would have to add the two EPE's at that stopping point.

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