# Physics QWatch

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#1
Hey guys, I'm struggling to understand Q3aii in the Jan 2012 paper for AQA found here:

http://filestore.aqa.org.uk/subjects...W-QP-JAN12.PDF

Basically, the mark scheme says you can just equate the girl's weight to kdelta L so essentially mg = Hooke's law = kdeltaL but I can't understand why the equation shouldn't be equal to ma. I understand why mg = kdeltaL if the rope was not accelerating but since it is then I can't seem to fathom out why the above is the case. Surely she is not accelerating at gms^-2 as she will be feeling the 'tension' from the rope which will reduce this value? It's been 2 years since I studied Hooke's Law so maybe I'm forgetting something fundamental here.

0
4 years ago
#2
(Original post by Saxandthecity)
Hey guys, I'm struggling to understand Q3aii in the Jan 2012 paper for AQA found here:

http://filestore.aqa.org.uk/subjects...W-QP-JAN12.PDF

Basically, the mark scheme says you can just equate the girl's weight to kdelta L so essentially mg = Hooke's law = kdeltaL but I can't understand why the equation shouldn't be equal to ma. I understand why mg = kdeltaL if the rope was not accelerating but since it is then I can't seem to fathom out why the above is the case. Surely she is not accelerating at gms^-2 as she will be feeling the 'tension' from the rope which will reduce this value? It's been 2 years since I studied Hooke's Law so maybe I'm forgetting something fundamental here.

At Q, the acceleration is zero as Q is the equilibrium point of the oscillating motion. If acceleration is zero, then the net force acting on the student should also be zero, thus .
1
#3
(Original post by NDVA)
At Q, the acceleration is zero as Q is the equilibrium point of the oscillating motion. If acceleration is zero, then the net force acting on the student should also be zero, thus .
So the F in the Hooke's law equation is not net force like the 'F' in Newton's Second Law, but a kind of tension from the spring? So typically if you have a moving object on a spring that is accelerating the force from Hooke's equation will just be a component of the overall net force?

Hope that makes sense, I'm half asleep today. Thanks for the help.
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