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That’s the definition of the acceleration due to gravity. What else could it be?
Original post by kev23s
Two masses hang at rest from a spring, as shown in the diagram. The string separating the masses is
burned through.

see the rest question below. why is acceleration of 2kg mass downwards is 1g ???? Q5


https://pmt.physicsandmathstutor.com/download/Physics/A-level/Topic-Qs/AQA/04-Mechanics-and-Materials/Set-M/Materials%20(Multiple%20Choice)%20QP.pdf

The link appears to be for a different question.
Original post by Stonebridge
The link appears to be for a different question.

Did you go down to Q5? I found the question (I think).
Original post by e^(i*pi) +
Did you go down to Q5? I found the question (I think).

Ha - I didn't scroll down. :colondollar:
Yes it's Q5. :smile:
Reply 5
Original post by e^(i*pi) +
That’s the definition of the acceleration due to gravity. What else could it be?

that's so true cant believe i forgot about that
The answer is C
How would you work out the upwards acceleration on this question? @Stonebridge
(edited 3 years ago)
Original post by AlanRoskilly
How would you work out the upwards acceleration on this question? @Stonebridge

At the start in equilibrium there is a total force of 3mg downwards. (The sum of the weights of the two masses.)
So the force of tension in the spring must balance that upwards.
Spring is applying an upwards force of 3mg.
When you cut the string, the remaining 1kg mass has its weight downwards of 1mg.
The spring is pulling it up with a force of 3mg.
Resultant upwards force is 2mg This acts upwards on the 1mg mass
This gives an acceleration upwards of 2g
Original post by tadpole245
Would the 2 kg mass accelerate downwards at 1g as Resultant Force=2g-g but for the 2kg mass you'd do the method you explained?


The 2-kg-mass will accelerate downward at 1g NOT because the resultant force = 2g-g.

When the string is cut, the 2-kg-mass experiences a resultant force of its weight, 2mg.
Using Fnet = ma, the 2-kg-mass will accelerate at 1g downward.

As this is a very old thread, we will close this soon and if you have any other queries, start a new thread and link to this thread. Thanks.

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