I was reading the definition of work done in terms of a kinetic energy.
It read that when a force is applied to a particle moving with constant velocity v, its kinetic energy changes as follows:-
$$\frac{dK}{dt} = \frac{1}{2}\frac{d(m\vec v.\vec v)}{dt} = \frac{1}{2}\frac{d(mv^2)}{dt} = m\vec v\frac{d\vec v}{dt}$$
The speed of a particle does not change when the force applied on it is perpendicular to the velocity and as a result, the kinetic energy of the particle also doesn't change.
Hence the force acting on the particle must have a tangential component.
Therefore:-
$$\frac{dK}{dt} = \vec F.\vec v$$
My query:
I know that the symbol $\vec v$ represents velocity(which is a vector) and $v^2$ or $\vec v.\vec v$ is a scalar. But then why is $\frac{d\vec v}{dt}$ equal to zero when the force applied on a particle is perpendicular since the velocity clearly is changing,at least its direction is.
Best Answer
The last expression should be a dot product:
$$\frac{dK}{dt} = m\vec v \cdot \frac{d\vec v}{dt}$$
So when a force is applied perpendicularly to the direction of motion, $\frac{d\vec v}{dt}$ is a vector that is perpendicular to $\vec v$, so their dot product is zero, giving you zero change in the kinetic energy as expected.