# Kinetic and Potential Energy

The mechanical energy of a system is the sum of its kinetic energy and potential energy.

## Kinetic energy

Kinetic energy is the energy that an object possesses due to its motion. Kinetic energy of a marticle of mass $m$ moving with a velocity $v$ is given by \begin{align} K=\frac{1}{2}mv^2. \end{align} The unit of kinetic energy is Joule. Its dimensions are ML2T-2. Kinetic energy K$and linear momentum$p$of a particle of mass$mare related by \begin{align} K=\frac{p^2}{2m}. \end{align} The work-energy theorem relates work done by conservative forces with the change in kinetic energy. ## Potential energy Potential energy is the energy that an object possesses due to its position or configuration. It is the energy stored in an object due to its position or configuration. The gravitational potential energy and elastic potential energy are two important forms of potential energy. ### Gravitational potential energy Gravitational potential energy is the energy an object possesses due to its position in a gravitational field. The gravitational potential energy of an object of massm$positioned at a height$h(above some reference) is given by \begin{align} U_g = mgh \end{align} whereg$is the acceleration due to gravity. ### Elastic potential energy Elastic potential energy is the energy stored in an object due to its configuration. The elastic potential energy of a stretched spring of spring constant$kis given by \begin{align} U_s = \frac{1}{2}kx^2 \end{align} wherex$is the difference between stretched and natural (equilibrium) lengths of the spring. ## Problems from IIT JEE Problem (JEE Mains 2022): A body of mass 8 kg and another of mass 2 kg are moving with equal kinetic energy. The ratio of their respective momenta will be 1. 1:1 2. 2:1 3. 1:4 4. 4:1 Problem (JEE Mains 2003): A spring of spring constant$5\times10^3\$ N/m is stretched initially by 5 cm from the unstretched position. Then the work required to stretch it further by another 5 cm is

1. 12.50 Nm
2. 18.75 Nm
3. 25.00 Nm
4. 6.25 Nm

Solution: The work required is equal to the change in spring potential energy, \begin{align} W&=U_f-U_i\\ &=\frac{1}{2}kx_f^2-\tfrac{1}{2}kx_i^2\nonumber\\ &=\frac{1}{2}(5\times{10}^{3}) (0.10)^2-\frac{1}{2}(5\times{10}^{3}) (0.05)^2 \nonumber\\ &=18.75\;\mathrm{J}.\nonumber \end{align}