Energy changes of a stretched string

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The discussion centers on the relationship between stretching a string and the resulting energy changes, particularly how a reduction in force leads to an increase in velocity and height. Participants explore the implications of Hooke's Law and the conservation of energy, questioning how a decrease in force can still result in increased kinetic energy as gravitational potential energy decreases. It is established that when the mass is below a certain point (R), the spring force exceeds gravitational force, resulting in positive acceleration and increasing velocity. The upward acceleration is identified as the reason for the height increase, aligning with the mark scheme's explanation. Overall, the conversation clarifies the dynamics of energy transfer in a stretched string system.
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Homework Statement


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This is the answer :

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For part d) of this question, I don't understand "stretching/extension reduces and velocity increases/height increases" .

Homework Equations


Hooke's Law: F=kx

The Attempt at a Solution


Ok, so if extension is reduce, then force reduces too. Then how does a decrease in force causes the velocity and height to increase? Does it have something to do with the principle of conservation of energy, where a gain in kinetic energy equals to a lose in gravitational potential energy?
 
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At a point lower than R , how does spring force compare to force of gravity ?
 
Qwertywerty said:
At a point lower than R , how does spring force compare to force of gravity ?

Spring force is more than the force of gravity .
 
Janiceleong26 said:
Spring force is more than the force of gravity .
So as long as mass is below R , acceleration is ( +ve or -ve ) ? And therefore velocity will always what ?
 
Qwertywerty said:
So as long as mass is below R , acceleration is ( +ve or -ve ) ? And therefore velocity will always what ?

Positive..? I'm not sure.. Therefore, velocity will increase?
 
Janiceleong26 said:
Positive..? I'm not sure.. Therefore, velocity will increase?
Yes . ( kx - mg = ma ) , kx > mg .

Now , a = dv/dt . As a is always positive till before R , dv is always +ve , and hence velocity will increase till R .

Hope this helps .
 
Qwertywerty said:
Yes . ( kx - mg = ma ) , kx > mg .

Now , a = dv/dt . As a is always positive till before R , dv is always +ve , and hence velocity will increase till R .

Hope this helps .

But how do you know that a is always +ve till before R? And why height increases? Does it got to do with 1/2 mv^2=mgh (gain in kinetic energy=lost in gravitational potential energy) ? And thanks by the way. :)
 
Janiceleong26 said:
But how do you know that the a is always +ve till before R?
I thought we agreed that -
Qwertywerty said:
( kx - mg = ma ) , kx > mg .
Janiceleong26 said:
And why height increases?
It rises because acceleration is upwards . And that's pretty much the only reason .
 
Qwertywerty said:
I thought we agreed that -It rises because acceleration is upwards . And that's pretty much the only reason .

Yeah, I agreed to that one. But the mark scheme also states that the height increases as well. Why?
 
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Janiceleong26 said:
Yeah, I agreed to that one. But the mark scheme also states that the height increases as well. Why?
Qwertywerty said:
It rises because acceleration is upwards . And that's pretty much the only reason .
 
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  • #11
OK thanks
 
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