# Does this vector need to be fixed for the kinetic energy to be constant?

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!

• 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. May 25, 2021 at 5:03
• 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)
– Eli
May 25, 2021 at 10:10

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.