Energy changes of a stretched string

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Janiceleong26
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Homework Statement


Exam question page showing a spring hanging vertically from a support with mass M attached at its lower end, labelled R and S for the extension positions, with part (a) on Hooke's law and part (c) on acceleration.
Exam question part (d) from a printed past paper, numbered 11, asking to describe and explain the energy changes from release of object M in Fig. 6.1 until it first returns to R, with answer lines and a [2] mark allocation.


This is the answer :

Mark scheme page from Cambridge AS/A Level physics paper 9702/22, October/November 2012, showing answers to question 6 parts (a) to (d) on spring extension, Hooke's law and energy changes.


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?
 
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?
 
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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