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Distinguish between potential energy and kinetic energy of a body.

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Hint:Energy is defined as the capacity to do any work. Whenever the object is at rest, the energy it contains is called the Potential energy. The Potential energy depends on the mass of the object and the height at which the object is placed. As soon as the object moves or comes into motion;

the Potential energy gets converted into Kinetic energy. The Kinetic energy depends on the mass of the object and the velocity of the object.

Complete step by step solution: The capacity to do work is called energy. Energy is a scalar quantity.

There are two types of energies. They are Kinetic Energy and Potential energy.
1. Kinetic Energy: The energy possessed by virtue of motion is called Kinetic Energy. If a body of mass \[m\] is moving with velocity \[v\], then its Kinetic energy is given by

\[K = \dfrac{1}{2}m{v^2}\]

When a body moves along a circular path with constant speed, its Kinetic energy remains constant.

kinetic energy of a body cannot change if the force acting on a body is perpendicular to the
instantaneous velocity. It can also be written as \[K = \dfrac{{{p^2}}}{{2m}}\], where \[p\] is the momentum and \[M\] is the mass. The kinetic energy is always positive.

2. Potential Energy: The energy possessed by a system due to its position or configuration is called potential energy. If a body of mass \[m\]is located at height \[h\] and if \[g\]is the acceleration due to gravity then, the potential energy of that body is given by
\[P = mgh\]

The potential energy is associated with mass in the gravitational field, compressed spring, deformed elastic body or charged capacitor. The potential energy can be positive or negative.

Note:According to the law of conservation of energy, the total energy of the body, that is the sum of kinetic energy and potential energy is conserved. The simplest example of conservation of energy is a vertically falling object. It has potential energy as it is at a height under the influence of gravity and has
kinetic energy as it changes its height. Therefore,
\[\

K + P = const \\
\dfrac{1}{2}m{v^2} + mgh = const \\
\ \]