Work-energy theorem:
Let us consider a particle is acted upon by a force \( \vec{F} \), as a result the particle is displaced by a distance \( d\vec{r} \). Then the work done by the force on the particle is given by,
\( dW=\vec{F}\cdot{d\vec{r}}\)
Since the work done is the scalar product of two vectors force and displacement, so the work done is a scalar quantity.
Now the total work done in moving the particle from the point \( A \) to point \( B \) along the curve \( C \) by the force \( \vec{F} \) is given by,
\( W=\displaystyle{\int_c}\left(\vec{F}\cdot{d\vec{r}}\right) \)
\( or,\ W=\displaystyle{\int_A^B}\left(\vec{F}\cdot{d\vec{r}}\right) \)
Now if the position vector of the points \( A \) and \( B \) are \( \vec{r_1} \) and \( \vec{r_2} \) respectively, then the work done by the force \( \vec{F} \) will be,
\( W=\displaystyle{\int_{r_1}^{r_2} }\left(\vec{F}\cdot{d\vec{r}}\right) \)
Let us consider, the particle has constant mass \( m \). At time \( t_1 \) the particle is at point \( A \) with velocity \( \vec{v_1}=\frac{d\vec{r_1}}{dt}=\vec{\dot{r_1}} \)
and, at time \( t_2 \) the particle is at point \( B \) with velocity \( \vec{v_2}=\frac{d\vec{r_2}}{dt}=\vec{\dot{r_2}} \)
Then the total work done by the force \( \vec{F} \) is given by,
\( W=\displaystyle{\int_{r_1}^{r_2} }\left(\vec{F}\cdot{d\vec{r}}\right) \)
\( or,\ W=\displaystyle{\int_{r_1}^{r_2}\left(\vec{F}\cdot\frac{d\vec{r}}{dt}dt\right)} \)
\( or,\ W=\displaystyle{\int_{r_1}^{r_2}\left(\vec{F}\cdot\frac{d\vec{r}}{dt}dt\right)} \)
\( or,\ W=\displaystyle{\int_{r_1}^{r_2}\left(m\frac{d\vec{v}}{dt}\cdot{\vec{v}}dt\right)} \)
\( or,\ W=m\displaystyle{\int_{r_1}^{r_2}}(\vec{v}\cdot{d\vec{v}}) \)
\( or,\ W=\displaystyle{\frac{m}{2}\int_{v_1}^{v_2}}d(\vec{v}\cdot\vec{v}) \)
\( or,\ W=\displaystyle{\frac{1}{2}mv^2|_{v_1}^{v_2}} \)
\( or,\ W=\displaystyle{\frac{1}{2}mv_2^2-\frac{1}{2}mv_1^2} \)
Thus the work done by the force \( \vec{F} \) during the time interval \( (t_2-t_1) \) to displace the particle of mass \( m \) through he displacement \( (\vec{r_2}-\vec{r_1}) \) is equal to the change in kinetic energy of the particle. This is the general statement of work-energy theorem.
When the speed of the particle is constant, i.e. the initial velocity \( \vec{v_1} \) is equal to the final velocity \( \vec{v_2} \), then the work done by the resultant force becomes zero.
Again, when the direction of displacement of the particle is at right angle to the direction of the force acting on the particle, then the work done by the force becomes zero, because \( dW=\vec{F}\cdot{d\vec{r}}=F\ r\ \cos\theta=F\ r\ \cos90^{\circ}=0 \).
For example, when a particle revolves in a uniform circular motion, then the centripetal force acting on the particle and the velocity of the particle are perpendicular to each other, for this reason, the centripetal force does not do any work done on the particle.
Kinetic energy of a particle is possessed by virtue of the motion of the particle. This kinetic energy is mechanical energy and it is measured by the amount of work done by the particle, before coming to rest, against the opposing force which tries to bring back the particle to rest.