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What is the unit of potential energy?

Can we explain charging by friction

How wave energy propagates in the direction of towards the wave front?

Give examples of kinetic energy and potential energy

Which of the following statement is true?
A. The energy stored in an object due to the position of that object is called kinetic energy.
B. The energy possessed by an object by virtue of its motion is called potential energy.
C. The energy stored in an object due to its position is called potential energy.
D. The energy possessed by an object by virtue of its motion is called kinetic energy.

Why is Kinetic Energy a Scalar Quantity?

If E= hv, find the unit of h .

Find the number of zeros after decimal point in ${0.2}^{25}$,given that $\log <!--\; \; -->2=0.30101$
My attempt: I found the answer as 17.…
Should we add 1 as 17 is the characteristic or leave it as 17?

What's U in Physics?

Evaluate ${\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}\frac{{\sin }^2<!--\; \; -->\left(t\right)}{t^2}dt$ with help of Laplace transform
Using the following identity
${\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}\frac{f\left(t\right)}{t}dt={\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}\mathcal{L}\left\{f\left(t\right)\right\}\left(u\right)du$
I rewrote
${\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}\frac{{\sin }^2<!--\; \; -->\left(t\right)}{t^2}dt$
as
${\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}\frac{{\sin }^2<!--\; \; -->\left(t\right)}{t\cdot <!--\; \cdot \; -->t}dt$
And thus the initial integral should be easily evaluated as
${\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}\mathcal{L}\left\{\frac{{\sin }^2<!--\; \; -->t}{t}\right\}\left(u\right)du$
According to my calculations, this is equal to
${\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}(\frac{1}{4}\ln <!--\; \; -->(u^2+4)-<!--\; -\; -->\frac{u^2}{4})du$
Which evaluates to $\frac{\mathrm{\pi <!--\; \pi \; -->}}{16}$. Being that this acutally a well-known integral and that its value is actually $\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}$ I think that I made a crucial mistake somewhere. Any ideas?

What is the difference between friction and inertia? Are they same in some aspect?

Consider a spring with spring constant k attached at one end to a wall. When an external force stretches it by distance l in the positive direction the force does work $\frac{1}{2}k{l}^2$ and that energy is stored in the spring as elastic potential energy.
Let the stretching take place uniformly over a time t for definiteness.
If this operation is viewed by an observer moving at velocity v with respect to the lab, the starting and ending position of the stretching operation will appear to be different, say 0 and $l-<!--\; -\; -->vt$, but the average force exerted on the spring would remain $\frac{1}{2}kl$, which seems to make the work done (and energy stored in the spring)
$W=\left(\frac{1}{2}kl\right)\cdot <!--\; \cdot \; -->(l-<!--\; -\; -->vt).$
How is this (potentially vast) difference reconciled?

Consider a spring with potential energy $U=\frac{1}{2}k{x}^2$, where x is displacement from equilibrium.
Hooke's law says the force required to move the spring is $F=kx$
From classical mechanics, we usually calculate force as $F=-<!--\; -\; -->\mathrm{\nabla <!--\; \nabla \; -->}U$ which yields $F=-<!--\; -\; -->kx$ here.
Why is the sign different? Does $F=-<!--\; -\; -->\mathrm{\nabla <!--\; \nabla \; -->}U$ denote the force on the spring instead of the force generated by the spring?

A spring with N turns radius R and spring constant k with initial length L. Current I is flown through it. Find the amount of compression in the spring.
I first tried this question using conserving energy and taking spring as an inductor. Initial magnetic energy = final magnetic energy + spring potential energy. But that didn't workout. Later I derived force on lateral surface of spring by using magnetic pressure but I don't know how to proceed.

The situation: I have an ideal spring, at equilibrium, sitting on a table. I then put a weight onto the spring which compresses the spring.
The spring has a height of 1m and spring constant k. The weight has a mass, m. The weight compresses the spring by 0.5m. If we take the table height to be the zero of gravitational potential energy, then the work done on to the spring, by the weight, is:
$W=-<!--\; -\; -->F\cdot <!--\; \cdot \; -->\mathrm{\Delta <!--\; \Delta \; -->}h=mg(0.5\text{m}-<!--\; -\; -->1\text{m})=\frac{1}{2}mg$
My question is: What is the work done on to the weight by the spring?
If we keep the conventions the same (spring is system, weight is surroundings), then the spring should be doing negative resistive work on to the weight while it's lowering.
In other words, shouldn't the work actually b

I am doing an experiment where I extended a spring from a equilibrium a set distance and then launch a ball with it. Then I see how high it goes and what its average speed is (total distance it flies divided by time).
The machine that has the spring in it is set at an angle of 45 degrees. The aim of the experiment is to determine a relationship between distance from equilibrium of the spring to the height and velocity of the balls motion (2 different relationships).
The formula that express what happens could be given as, $0.5k{x}^2=mgh+0.5m{v}^2.$
My question is how much of the spring potential energy will be converted into gravitational energy and how much will be converted into kinetic energy (both at the highest point of the motion).

Suppose a mass of $M$ kg is hanging from a spring in earth. The mass will stretch the spring about x m. So the change in the gravitational potential energy is $mgx$ J (supposing x to be very small compared to the radius of earth).
And this amount of energy will be stored in the spring as potential energy. So,
Change of gavitational energy = $mgx$ = potential energy stored in the spring
And it seems that the potential energy stored in a spring is proportional to displacement x. But the potential energy in a spring is $U=\frac{1}{2}k{x}^2$ and so it's proportional to $x^2$, the square of displacement. So surely I am wrong somewhere. But where am I wrong?

I've been given that there is a spring holding a $.25kg$ mass where $k=10N/m$ and is held at $40cm$ and then let go. I've found that the max velocity is $2.53m/s$ and that when the spring is at $20cm$ the velocity is $2.19m/s$What I'm having a hard time wrapping my head around is finding the position when the kinetic energy and potential energy are equal.
I keep trying to arrange it like so but end up with two unknowns
$.5(10N/m)(\mathrm{\Delta <!--\; \Delta \; -->}x{)}^2=.5(.25kg)v^2$

Kinetic energy in a spring
I am building a spring launcher for an engineering project and need to calculate spring compression to achieve a projectile velocity of 4 m/s. The mass $m_p$ of the projectile is 5 g, the mass of the spring is 7 g and the spring constant is 1000 N/m.
Is it right to assume that initial potential energy in the spring = kinetic energy of the marble + kinetic energy of the spring at the point of release?
I can calculate the marble's kinetic energy easily enough using $E_k=\frac{1}{2}{m}_p{v}^2$. However, how would I calculate the spring's kinetic energy? One end is secured, while the other end would be traveling at the same speed as the marble. I think I would have to use integration somehow but not completely sure how to.

We know that a stretched spring obeys Hooke's law, such that $F=-<!--\; -\; -->kx$
We can find the potential energy of stretching/compressing this spring by x, given by :
$U_x-<!--\; -\; -->U_0=-<!--\; -\; -->{\int <!--\; \int \; -->}_0^{x}F.dx=\frac{1}{2}k{x}^2$
Setting $U_0=0$ as reference, we have $U_x=\frac{1}{2}k{x}^2$
However, this is also sometimes described as the work done by the spring.
Shouldn't the work done W be given by $\int <!--\; \int \; -->F.dr$, such that $W=-<!--\; -\; -->\mathrm{\Delta <!--\; \Delta \; -->}U=-<!--\; -\; -->\frac{1}{2}k{x}^2$ in this case ?
Isn't the work done by the spring negative ?
Also, in this case the potential energy comes to be negative.. In general, can we set any point as reference and set it to be 0 and perform the integral between any two limits, to get either a positive or a negative U ?
For example, in forces of the nature $r^{-<!--\; -\; -->n},(n>1)$ we usually take the reference at $r=\mathrm{\infty <!--\; \infty \; -->}$ and integrate from $\mathrm{\infty <!--\; \infty \; -->}$ to some point r. In case of forces of the nature $r^n$, we usually take 0 as the reference and integrate from 0 to some r. In general, we are free to choose any reference and any limit, even though some are much more convenient, right ? In theory, we can choose any point, right ?
As long as we have :
$U_a-<!--\; -\; -->U_b=-<!--\; -\; -->{\int <!--\; \int \; -->}_b^{a}F.dx$
we can choose any a and b, and set either of $U_a$ or $U_b$ to be the reference and equal to 0, right ?