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You have a mass moving under the influence of a simple massless spring. The equations of motion are $F = -k x = m a$. These can be solved using direct integration. \begin{aligned} a =\frac{{\rm d}v}{{\rm d}t} & = -\frac{k}{m} x \\ \int \frac{{\rm d}v}{{\rm d}t} \, {\rm d}x & = - \int \frac{k}{m} x \,{\rm d} x \\ \int v \, {\rm d}v ... 1 There are a few problems with your derivation. Your main mistake is using formulae that assume the acceleration of the spring is constant (it is not). In one dimensional systems with nonuniform acceleration, we have v(t) = v_0 + \int_{t_0}^t a(t)dt . $$Furthermore, the v^2 = v_0^2 + 2a \Delta x no longer works for nonuniform accelerations either. ... 2 One can't discuss the difference between factors of 1/3 and 1/6 without mathematics. The difference between these two numbers – and generally, any fact about any numbers – is all about mathematics. If all values of x between 0 and A were equally likely, the average value of kx^2/2 would be kA^2/6 as you say because the average value of X^2 ... 1 The spring would extend at a below light speed (but very close to it) this is because of it having some mass as you referred the spring to being "nearly mass-less" but it would extend on each side, therefore no physical law is broken however relative to an observer it would seem as if the object is expanding at faster than c (Speed of Light) speed however ... 3 You may be imagining that if you push with constant force F, the spring will compress until the spring has such a resistive force. But since the spring was not counteracting that force, your constant force F was accelerating the mass. Upon reaching the point where the spring has force F as well, the mass does not stop but has a speed such that KE = ... 0 The factor \frac{1}{2} is due to the integral. The wrong sign of yours is due to the fact that you have to counter the force of the spring. So the Force if the Spring is -kx, but you have to pull in the direction it is extended, so apply the force kx, therefore the energy is positive W=\frac 1 2 k^2 x 5 Since the force is a function of distance, you need to integrate:$$F = kx\\ W = \int F\ dx\\ W = \int k\ x\ dx\\ W = \frac12kx^2 Add signs as needed... Your work considered the force to be constant - and that's not how springs work.

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To get the initial conditions you guess and look at whether the guesses fit the physical situation. For example, suppose that you have a mass on a spring and you are holding mass so that the spring is slightly stretched and then release it. The mass is not moving at t=0 so the initial condition is that the velocity is zero at t=0. But you could imagine other ...

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I know which book you are referring to. It is the book "Finite elements in Engineering" by Chandrupatla and Balegundu.I also have the same question. We learnt in Physics that the Work done by the force is stored as Potential Energy. There was no mention of Work Potential.

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Solution 1---Guess-work: If the forces applied on the two ends are equal, say both $1.5\ N$, the spring will get stretched $1.5\ m$. A natural guess is that the stretch is determined by the average of the two forces at the ends, which in this case are both equal to $1.5\ N$. Therefore, for the case you mentioned ($1\ N$ applied to one end, and $2\ N$ to the ...

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