Kinetic Energy;

The kinetic energy of an object is the energy that the object possesses by virtue of its motion.

For example, when a bullet is fired from a revolver, then the bullet can penetrate a target due to its kinetic energy.

Expression for kinetic energy:

Let us consider an object of mass \( m \) be at rest, i.e. the initial velocity of the object is \( \vec{u}=0 \). Now due to the application on an external force \( \vec{F} \), the object starts moving with the velocity \( \vec{v} \) along the direction of the applied force.

Let \( \vec{r} \) be the displacement of the object during the time interval \( t \).

So work done by the force acting on the object is

\( W=\vec{F}\cdot\vec{r}=Fr\ \cos{0^{\circ}}=Fr\tag{i} \)

Again, \( \vec{F}=m\vec{a} \), where \( \vec{a} \) be the aceleration of the object.

Putting the value of \( \vec{F} \) is the equation (i) we get,

\( W=m\ a\ r \)

Again we know that,

\( v^2=u^2+2ar=0+2ar=2ar \)

\( or,\ \displaystyle{a=\frac{v^2}{2r}} \)

Therefore the work done is given by,

\( \displaystyle{W=m\frac{v^2}{2r}r} \)

\( or,\ \displaystyle{W=\frac{1}{2}mv^2} \)

Work done is equal to the kinetic energy of the object, beacuse work done on the object by the force sets the object in motion. Therefore,

\( K.E. = \displaystyle{\frac{1}{2}mv^2} \)

Work done is the change in kinetic energy:

Let us consider, a particle of constant mass \( m \) is moving under the influence of an external force \( \vec{F} \).

Let, \( d\vec{r} \) be the displacement of the particle due to the application of the external force \( \vec{F} \) on the particle, then the work done on the particle by the force is given by,

\( dW=\vec{F}\cdot{d}\vec{r} \)

since only the component of \( \vec{F} \) along the direction of \( d\vec{r} \) is responsible for the movement of the particle.

Let \( \vec{v_1} \) be the velocity of the particle at the initial point A at time \( t_1 \) and \( \vec{v_2} \) be the velocity of the particle at the final point B at time \( t_2 \).

So the work done by the external force \( \vec{F} \) in moving the particle from point A to point B along the path C is given by,

\( W=\displaystyle{\int_{C}\vec{F}\cdot{d}\vec{r}} \)

\( or,\ \displaystyle{W=\int_{t_1}^{t_2}\vec{F}\cdot\frac{d\vec{r}}{dt}dt} \)

\( or,\ \displaystyle{W=\int_{t_1}^{t_2}\vec{F}\cdot\vec{v}dt} \)

\( or,\ \displaystyle{W=\int_{t_1}^{t_2}m\frac{d\vec{v}}{dt}\cdot\vec{v}dt} \)

\( or,\ \displaystyle{W=\int_{t_1}^{t_2}m\vec{v}\cdot{d}\vec{v}} \)

\( or,\ \displaystyle{W=\frac{1}{2}m\int_{t_1}^{t_2}d(\vec{v}\cdot\vec{v})} \)

\( or,\ \displaystyle{W=\frac{1}{2}mv^2|_{v_1}^{v_2}} \)

\( or,\ \displaystyle{W=\frac{1}{2}mv_2^2-\frac{1}{2}mv_1^2 }\)

So total work done in moving the particle along the path C from A to B is

\( \displaystyle{W=\frac{1}{2}mv_2^2-\frac{1}{2}mv_1^2} \)

\( or,\ W=T_2-T_1 \)

where, \( \displaystyle{T_1=\frac{1}{2}mv_1^2} \) is the kinetic energy at point A

and, \( \displaystyle{T_2=\frac{1}{2}mv_2^2} \) is the kinetic energy at point B.