I was solving the following homework problem:

A force $\vec{F} = \vec{k} \times \vec{v}$ is applied to a particle of mass $m$. Here $\vec{k}$ is a fixed vector and $\vec{v}$ is the velocity of the particle. Show that the kinetic energy of the particle is constant.

My attempt at a solution was as follows. We know that the work done by a particle being subjected to some force $\vec{F}$ from $a$ to $b$ is the change in kinetic energy. Using this, if I show that the work is $0$ for any starting time $a$ and ending time $b$, then the kinetic energy must be constant. We thus see that $$ W_{ab} = \int_{a}^{b} \vec{F} \cdot \vec{v} \ dt = \int_{a}^{b} (\vec{k} \times \vec{v})\cdot \vec{v} \ dt = \int_{a}^{b} \underbrace{(\vec{v} \times \vec{v})}_{\color{purple}{\vec{0}}}\cdot \vec{k} \ dt = 0 $$ and since $a$ and $b$ are arbitrary, the kinetic energy is constant.

While thinking about this solution, I noticed that I didn't use the fact that the vector $\vec{k}$ was fixed. So does this mean that the kinetic energy is still constant even if $\vec{k}$ isn't constant? Or did I make a mistake in my analysis? Thank you very much!

  • 1
    $\begingroup$ Your conclusion is correct, if the vector k where not constant, it would need to be cast as a path dependent quantity in the integrand, but the argument would still hold. Adding to this, the force used in this exercise is analogous to the magnetic field force on a charged particle, so this tells you the magnetic field doesnt do work on charged particles. $\endgroup$ May 25, 2021 at 5:03
  • $\begingroup$ from Newton law $ m\dfrac{d^{2}x}{dt^{2}}=F $,you obtain this equation $\dfrac{mv^{2}}{2}=\int \overrightarrow{F}\cdot \overrightarrow{v}dt=0$ thus v is zero or constant ( Initial condition) $\endgroup$
    – Eli
    May 25, 2021 at 10:10

1 Answer 1


The work done is zero because the force is perpendicular to the velocity, it will be zero even if $\vec{k}$ is time-dependent. Think of it like this: Work done by a Lorentz force is zero even if the magnetic field is time-dependent.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.