# Conservation of Energy , Significance of potential energy

Conservation of energy means conservation of all forms of energy together. Accounting all forms of energy within an isolated system the total energy remains constant.
While mechanical energy accounts only two forms of energy namely kinetic energy K and potential energy U.

If only conservative forces act on a system then total mechanical energy of the system remains constant.

i.e. K + U = Constant   ……..(1)

Therefore, ΔK + Δ U = 0   …………(2)

i.e. ΔK = – Δ U   ………(3)

This can be understood with reference to an example where, a block of mass M is attached to one end of a light spring of force constant K as, shown in figure . The other end of the spring is fixed to a vertical wall. The block is kept on a smooth horizontal surface. If the block is pulled outward and held at rest, the spring is elongated through a distance.

Now at this position the magnitude of work done by the spring force is numerically equal to
$\displaystyle \frac{1}{2}k x^2 = U(x)$ is the measure of potential energy and as the block is at rest the kinetic energy is zero, shown in figure .

Therefore total mechanical energy at this position is

$\displaystyle K + U(x) = 0 + \frac{1}{2}k x^2$ …(4)

Now the block is released and just when the spring comes back to its unstretched condition (figure ), then U(x) becomes zero. At that position let the velocity of the block be v0

Therefore, the total mechanical energy becomes kinetic energy,

$\displaystyle K + U(x) = \frac{1}{2}m v_0^2 + 0$ …. (5)

If we consider any intermediate position of stretching of the spring x then we will have the total mechanical energy, shown in figure , as

$\displaystyle K + U(x) = \frac{1}{2}m v^2 + \frac{1}{2}k x^2$ …. (6)

As the spring force is conservative, we have

$\displaystyle 0 + \frac{1}{2}k x^2$

$\displaystyle = \frac{1}{2}m v_0^2 + 0$

$\displaystyle = \frac{1}{2}m v^2 + \frac{1}{2}k x^2$ …….(7)

Thus under conservative force there is mutual exchange between kinetic and potential energy.

### Significance of potential energy:

We know in case of any conservative force, the potential energy is a function of its position.
U = U(x)

since total mechanical energy (K + U)is conserved.

$\displaystyle \frac{d}{dx}(K + U) = 0$

Therefore , $\displaystyle \frac{d}{dx}(\frac{1}{2}m v^2 + \frac{1}{2}k x^2) = 0$

$\displaystyle m v .\frac{dv}{dx} + k x = 0$  \displaystyle m a + k x = 0  \displaystyle m a = – k x  \displaystyle \frac{d}{dx}(K) = m a $and ,$ \displaystyle \frac{d}{dx}U(x) = – k x $Hence we can say that the negative of differentiation of U(x) with respect to x, gives the force acting on the system.$ \displaystyle F (x) = – \frac{d U(x)}{dx} $Illustration : The potential energy of a system of two particles is given by U(x) = a/x2 – b/x. Find the minimum potential energy of the system, where x is the distance of separation; a, b are constants. Sol:$ \displaystyle U(x) = \frac{a}{x^2} – \frac{b}{x}  \displaystyle F = – \frac{d U(x)}{dx}  \displaystyle F = – \frac{d}{dx} (\frac{a}{x^2} – \frac{b}{x} )  \displaystyle F = – (-\frac{2 a}{x^3} + \frac{b}{x^2} )  \displaystyle F = (\frac{2 a}{x^3} – \frac{b}{x^2} ) $When the particle is in equilibrium , F = 0$ \displaystyle (\frac{2 a}{x^3} – \frac{b}{x^2} ) = 0  \displaystyle x = \frac{2 a}{b} $Therefore, the minimum potential energy of the system is obtained by putting$ \displaystyle x = \frac{2 a}{b}  \displaystyle U(x) = \frac{a}{x^2} – \frac{b}{x}  \displaystyle U(min) = \frac{a}{(2a/b)^2} – \frac{b}{(2a/b)}  \displaystyle U(min) = -\frac{b^2}{4 a} \$