A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane. It follows that :

Step 1: Analyze the work done by the force.

Since the force $\vec{F}$ is always perpendicular to the velocity $\vec{v}$, their dot product is zero at every instant.

$$\vec{F} \cdot \vec{v} = 0$$

The work done $W$ by the force is the integral of power over time.

$$W = \int \vec{F} \cdot d\vec{r} = \int (\vec{F} \cdot \vec{v}) \, dt = 0$$

Step 2: Determine the kinetic energy and speed.

By the work-energy theorem, the net work done on the particle equals its change in kinetic energy.

$$\Delta K = W = 0$$ $$K = \frac{1}{2}mv^2 = \text{constant}$$

Since the kinetic energy is constant, the speed $v = |\vec{v}|$ of the particle is also constant.

Step 3: Analyze the velocity and acceleration vectors.

Since a perpendicular force acts on the particle, the direction of the velocity vector changes continuously.

$$\vec{v} \text{ is not constant}$$

The acceleration vector is given by Newton's second law.

$$\vec{a} = \frac{\vec{F}}{m}$$

Since the magnitude of the force $F$ is constant, the magnitude of the acceleration $a = \frac{F}{m}$ is also constant. However, because the direction of $\vec{F}$ changes to remain perpendicular to $\vec{v}$, the direction of $\vec{a}$ changes continuously.

$$\vec{a} \text{ is not constant}$$

Step 4: Determine the path of the particle.

For planar motion where the acceleration is always perpendicular to the velocity, the acceleration is purely centripetal.

$$a = \frac{v^2}{R}$$

Solving for the radius of curvature $R$:

$$R = \frac{v^2}{a}$$

Since both the speed $v$ and the magnitude of acceleration $a$ are constant, the radius of curvature $R$ is constant. A planar trajectory with a constant radius of curvature is a circle.

Answer: The kinetic energy of the particle is constant, and the particle moves in a circular path.