How Much Force to Apply on m1 to Make m2 Jump Off the Table?

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To make m2 jump off the table, the force applied on m1 must exceed the gravitational force acting on m2. The key equations involve Newton's second law and Hooke's law, indicating that the force F must be greater than the weight of m2 (F > m2 * g) at the moment of release. The confusion arises from the fact that m1 is already exerting a force on m2 due to its weight. Thus, the total force applied must account for both m1 and m2 to ensure m2 experiences a net upward force sufficient to overcome gravity. Understanding the net force at the instant of m2's jump is crucial for solving the problem.
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Homework Statement



Hey there,

We assume that the spring got no mass and there are no frictions. If you want to push down just as hard on m1 that if you release ... m2 will be just about to jump and leave the table?


unbenannt_979.jpg


Homework Equations



Newton and Hooke are our very best friends :)

The Attempt at a Solution



1. F > (m1 + m2) * g

or

2. F > m2 * g ?

Im a bit confused, as m1 is already pushing on the string even if there is no extra force... but how much do I have to push down for myself on m1? My guess is that 1 would be to total amount of force "pulling" on m2, but the force needed to get this right, would be 2, as m1 is supporting.

Thanks in advance
 
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What is the condition that m2 jumps and leaves the table? What can you say about the net force acting on it at that instant?

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