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Which strategy should I use to approach a problem? Sometimes I'll use $U=1/2k{x}^2$ when I should use $F=-<!--\; -\; -->kx$ and vice versa. What are clues in a problem that would let me identify the correct approach to a given problem? And shouldn't both approaches give me the same answer in any scenario?

Object attached to spring is at rest.
A force is then exerted in the positive x-direction that at each position is equal and opposite to the force exerted by the spring - in other words, a force is exerted such that the object on the spring moves with constant velocity in the positive x-direction.
Net work on object is equal to 0. So then where does the "extra" potential energy in the object come from, when it is displaced by that force in the positive x-direction? If I pull an object in this way 5m to the right and keep it at 5m, its potential energy has increased but kinetic energy hasn't.
Please explain.

If I were to stretch a spring, then I am doing positive work in order to increase the potential energy stored in the spring. Since the equation for the potential energy in this case is given by
$U(x)=\frac{1}{2}k{x}^2$
i.e, the potential energy depends only on the displacement from a reference point. Then would there be a conservative force doing negative work on the spring in order to increase the potential energy or not since $U\propto <!--\; \propto \; -->x^2$ rather than x?

If a spring is stretched by a weight of mass m, (so the extension is Δx) then
$k=\frac{W}{\mathrm{\Delta <!--\; \Delta \; -->}x}=\frac{mg}{\mathrm{\Delta <!--\; \Delta \; -->}x}$. So $k\mathrm{\Delta <!--\; \Delta \; -->}x=mg$
When the Spring is stretched by distance $\mathrm{\Delta <!--\; \Delta \; -->}x$ (by the weight) then it looses gravitational potential enegry. ($\mathrm{\Delta <!--\; \Delta \; -->}GPE=mg\mathrm{\Delta <!--\; \Delta \; -->}x$)
But, when we calculate the change in elastic potential energy, We get
$U=\frac{1}{2}k(\mathrm{\Delta <!--\; \Delta \; -->}x{)}^2$. Since $k\mathrm{\Delta <!--\; \Delta \; -->}x=mg$, $U=\frac{1}{2}mg\mathrm{\Delta <!--\; \Delta \; -->}x$
At, equilibrium we have no kinetic energy $E_k=0$
Doesn't that violate Energy Conservation?

According to my book the spring constant is given by $k=V^{″}(X_o)$. Where $V(X)$ is the potential energy function. if I use the function $V(X)=X^6+X^4$ the spring constant is zero at $X_o=0$. However, the plot looks like a stable equilibrium that would allow harmonic oscillation near $X_o$. So why is k zero in this case?

An object with the mass m is moving with velocity v. Its kinetic energy (k) is defined by $K=\frac{1}{2}m{v}^2$ and its momentum (p) by p=mv. Write its kinetic energy interms of its mass and momentum, but Not its velocity.

The formula for elastic potential energy is $(1/2)\mathrm{k}\mathrm{x}²$, all that I am asking is why is it not $\mathrm{k}\mathrm{x}²$. Here is my logic. Let us say that k is $2\mathrm{N}/\mathrm{m}$ which means that for the spring to move $1\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}$ you have to apply 2 N of force, if you want to move the spring $2\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}$, then you have to apply $2\mathrm{N}$ and then $2\mathrm{N}$ again, so the work should be $2\mathrm{N}\times 4\mathrm{m}=8\mathrm{J}$. Where did I go wrong?

Suppose we take a spring and compress it in a clamp. There is a potential energy due to the compression of the spring. It is now placed into a bath of acid that will dissolve the spring but not the clamp. What happens to the potential energy?
In a realistic situation one (weak) point on the spring could break, so it would then just ping out of the clamp and lose the energy that way. But in the ideal scenario where it dissolves at a uniform rate, this wouldn't happen, so what happens at the instant that it is completely dissolved?
I thought maybe some kind of change of temperature would occur, due to the assumption that the compression would cause the atoms to be more tightly compacted causing a repulsion. As you remove layers this would gradually decrease, but there is still an inherent increase in energy for each atom, so this would most likely be expelled as something (like) heat.
I'm mainly interested in the detail of the mechanisms for the energy tranfers on an atomic level as opposed to the broad picture. Specifically the transfer from coulomb repulsion to heat in the acid. The more low-level detail the better.

Is the work done by external constant force = to potential energy stored in the spring.
i.e. $Fx=\frac{1}{2}K{x}^2$
But we know $F=kx$ but then $Fx=K{x}^2$

I was wondering if potential energy for spring is always positive.
$PE=\frac{1}{2}k{x}^2$
where k is spring constant and x is the change in length.

According to my physics book, the spring constant can be calculated from knowing the potential energy, with the formula $k=W_p^{″}(0)$
I don't really understand why, and the book doesn't explain it any further. How do I know the two are equal?

We generate the formula from $W=Fs$, but why is it when we form the formula the $F$ becomes the average force applied, making the equation
$W=(1/2)(0+F)x?$

Why is the gravitational potential energy of ideal uniform massive spring $mgx/2$, not $mgx$?
$L=T-<!--\; -\; -->V=\frac{1}{2}\frac{m}{3}{\stackrel{\dot{}<!--\; \dot{}\; -->}{x}}^2+\frac{1}{2}M{\stackrel{\dot{}<!--\; \dot{}\; -->}{x}}^2-<!--\; -\; -->\frac{1}{2}k{x}^2-<!--\; -\; -->\frac{mgx}{2}-<!--\; -\; -->Mgx$
where $mgx/2$ refers to gravitational potential energy of ideal spring. But I do not get why it's $mgx/2$ instead of $mgx$. What is the reason?

A point mass m having charge q is connected with massless spring to a rigid wall on a horizontal surface.A horizontal uniform electric field E is switched on and mass is displaced through a distance x from equilibrium position.
Work done by electric force is $W=qE.x$ as electric force is constant.
But F(external) on spring mas system is equal to kx. So $W=kx.x=k{x}^2$ (equation 1) as external force i.e, electric force is constant.
We know that work done is stored in the form of potential energy and we know that potential energy stored in the spring is $1/2k{x}^2$.
So $W=1/2k{x}^2$ (let it be equation 2)
But the above 2 equations are not equal. From where did the 1/2 come from in the 2nd equation?

Is there a difference between the potential energy of an oscillator and the potential energy of the spring which is associated with? When we compress a spring horizontally without any object associated with it starts to expand? Is the potential energy convert into kinetic energy of the spring?

Let's consider having 2-D Cartesian coordinate xy-plane, a block connected to a spring (placed horizontally on negative part of x-axis and having natural length L) sitting at the origin point with the spring at natural length.It is allowed to oscillate back and forth about the origin. From the equation that relates the work done by conservative force and the potential energy, $U(x_a)-<!--\; -\; -->U(x_b)=-<!--\; -\; -->{\int <!--\; \int \; -->}_{x_b}^{x_a}{F}_x\text{d}x$. If we define the zero point at x=0, we can express its potential energy for $0<<!--\; <\; -->x<<!--\; <\; -->L$, with
$U(x)-<!--\; -\; -->U(x=0)=-<!--\; -\; -->{\int <!--\; \int \; -->}_0^x-<!--\; -\; -->kx\text{d}x\Rightarrow <!--\; \Rightarrow \; -->U(x)=\frac{1}{2}k{x}^2$
while for −L<x<0, we have (the restoring force switch sign as it now points to the positive x- direction):
$U(x)-<!--\; -\; -->U(x=0)=-<!--\; -\; -->{\int <!--\; \int \; -->}_0^{x}kx\text{d}x\Rightarrow <!--\; \Rightarrow \; -->U(x)=-<!--\; -\; -->\frac{1}{2}k{x}^2$
My question: Is my idea and this approach for the energy of the system correct? Because I saw from some reference books, they simply say that it is just $\frac{1}{2}k{x}^2$ for both of the conditions.

From classical mechanics, the force on a spring is given by the negative gradient of the potential energy with respect to position or displacement.
Can we also say $F=-<!--\; -\; -->\mathrm{\partial <!--\; \partial \; -->}U/\mathrm{\partial <!--\; \partial \; -->}x$, where U is the "internal" energy of the spring?

The total energy of an object comes from the time part of the four-momentum, and so isn't a Lorentz invariant. On the other hand, is the potential energy of a compressed spring a Lorentz invariant?

While working with problems on elastic collisions, I have come across this observation, that the elastic potential energy of a two-body system is the maximum when the relative velocity equals zero. In fact, the same applies when we are talking about a 2-block+spring system. Why is it so?

What is the electric potential at the point indicated with the dot in the figure? Express your answer to two significant figures and include the appropriate units.
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