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1 vote

It's more like a simple math question but how does this process to that?

$\frac{mg}{k}$ is a constant term not dependent on time and hence if it is differentiated (twice) with respect to time it becomes zero. So mathematically the two terms highlighted in yellow only ...
Farcher's user avatar
  • 97.9k
1 vote

What are the forces that do work when a spring between two masses pulls inwards?

There is no kinetic energy change in the second system (1,2, spring). So there is no net work done, so no need to ask which external forces that do the work. The constituent particles of the system do ...
Steeven's user avatar
  • 51.5k
7 votes

Spring potential energy, conversion

The problem in your book assumes an ideal (lossless) and massless spring. That is typical of entry level physics problems. Without the ball you would need to consider, at a minimum, that the spring ...
Bob D's user avatar
  • 73.7k
5 votes

Spring potential energy, conversion

In the case where there is no ball (and assuming there is no friction/other dissipative forces), the spring will expand beyond its natural length due to the momentum it gains while unstretching. At ...
CompassBearer's user avatar
1 vote

Strain energy stored in a bungee cord pulled at its midpoint

When a spring is halved, its spring constant $k$ is doubled. Because I'm treating this as a two spring system I need to double the value of $k$ that I'm using. Using a value of $k=280$ N/m gives $x=....
Imperator's user avatar
2 votes

What happens to the amplitude when a spring is compressed?

In the case of an ideal spring, this situation sets up a simple harmonic oscillator with an amplitude equal to the distance you compressed the spring by. Note that the amplitude being equal to that ...
Riley Scott Jacob's user avatar

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