Physics with Calculus/Mechanics/Energy and Conservation of Energy/Potential energy

Potential energy is the energy stored in an object due to its position. There are several types of potential energy.

Gravitational
Gravitational potential energy, involves the line integral of the force between two objects ($$m_1$$ and $$m_2$$). By Newton's universal law of gravity, the force is

$$\mathbf{F}_g = -\frac{Gm_1 m_2}{r^2}\;\hat r$$

We integrate to get potential energy:

$$U_g(r) = -\int_\infty^r \mathbf{F}_g\,dr' = \int_\infty^r \frac{Gm_1 m_2}{r^2}\,dr = -\frac{Gm_1 m_2}{r}$$

Here, we have taken the reference point (where the potential energy equals zero) to be at $$r=\infty$$. Sometimes, when dealing with small distances where the difference in acceleration due to gravity will be negligeable we simplify the energy equation by assuming that $$r = R +y$$, where $$R$$ is the Earth's radius and $$y<<R$$ is the height above the Earth's surface. Taking $$m_2$$ to be the mass of the planet:

$$F_g = \frac{Gm_1 m_2}{R^2}$$

$$g = \frac{Gm_2}{R^2}$$.

Note that the vector $$g$$ points in the $$-\hat r$$ direction. Inserting this into the integral for $$U_g$$:

$$U_g = -\int_0^y (-m_1g\hat r)dr'= m_1 gy$$,

where now, the reference point is on the surface of the Earth.

Elastic
Elastic potential energy is the energy stored in a compressed or elongated object (a spring, for example). The amount of energy stored in the object depends on spring constant ($$k$$) and the displacement from the rest position ($$x$$). It should be noted that the amount of energy is the same regardless whether the object is compressed or elongated. Given the force:

$$\mathbf{F}_s = -k\mathbf{x}$$

We integrate to get energy:

$$U_s = -\int \mathbf{F}_s\,dx = \int -k\mathbf{x}\,dx = \frac{1}{2}k\mathbf{x}^2$$