Question

What is the derivative of kinetic energy with respect to velocity?

Answer 1

Hint: We know that the kinetic energy of an object is the energy that it possesses due to its motion. It is defined as the work required to accelerate a body of a given mass m from rest to its accelerated state. We will find the derivative of kinetic energy with respect to velocity using basic formulas of differentiation.

Formula used:
Let $y=ax^n$, where $a$ is a constant, and $y$ changes according to $x$.
Derivative of $y$ with respect to $x$ is $\dfrac{dy}{dx}=a.n x^{n-1}$

Complete step by step solution:
We know that kinetic energy is the energy of a particle acquired due to the motion of the particle from a rest position. A particle at rest has potential energy stored in it and it is converted into kinetic energy when it starts moving or accelerating. Having gained this energy during its acceleration, the body maintains the same kinetic energy unless its speed changes.
And the kinetic energy of the particle will be given by
$K = \dfrac{1}{2}m{v^2}$
Now differentiating the equation with respect to velocity we get
$\dfrac{{dK}}{{dv}} = \dfrac{d}{{dv}}\left[ {\dfrac{1}{2}m{v^2}} \right]$
Since the mass of the particle constant therefore by linear property of derivative we get
$\dfrac{{dK}}{{dv}} = \dfrac{1}{2}m\dfrac{d}{{dv}}{v^2}$
We know that derivative of a power function, therefore we get
$\therefore \dfrac{{dK}}{{dv}} = \dfrac{2}{2}mv = mv = p$
We know that the linear momentum of a particle moving with a velocity \[v\] and mass $m$ is given by $p = mv$
Therefore the derivative of kinetic energy with respect to velocity is equal to the momentum of the particle.

Note: Note that the mass of the particle is constant in classical mechanics. If we take into account the relativistic effects we get the same result but the derivation becomes complicated. The relativistic mass becomes infinite as the velocity of the body approaches the speed of light.