Static equilibruim in light spring

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SUMMARY

The discussion focuses on the static equilibrium of a light spring with a mass of 0.20 kg and an additional mass of 0.10 kg suspended from it. When the thread connecting the two masses is burned, the upward acceleration of the 0.20 kg mass is calculated to be 5.0 m/s², based on the total weight acting on the spring being 3 N. The key to solving the problem lies in analyzing the forces acting on the upper mass just before the string is severed, as the spring's force remains unchanged at that instant.

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jinx007
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A light spring has a mass of 0.20 kg suspended from its lower end. A second mass of 0.10 kg is suspended from the first by a thread. The arrangement is allowed to come into static equilibrium and then is burned through. At this instant, what is the upward acceleration of the 0.20 kg mass. ( take g= 10 m s^-2)

Answer according to the booklet is 5.0 m s^-2

total weight acting on the first spring is 3 N ( i think so..)

Then i don't really know how to proceed...help..!
 
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You can do this : look at the force on the upper mass just before the string is burnt through since the spring continues to exert the same force as long as the extension does not change (which does not at that instant as the problem says), all you have to do is take out the tension of the now cut string and you will have the force and acc.
 

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