Animated Solution for Physics - Work, Energy, and Power: A particle of mass $0.2\text{ kg}$ is moving in one dimension under a force that delivers a constant power $0.5\text{ W}$ to the particle. If the initial speed (in ${\text{ms}}^{-1}$) of the particle is zero, the speed (in ${\text{ms}}^{-1}$) after $5\text{ s}$ is

The problem of a particle accelerating under constant power is a classic and beautiful application of the Work-Energy Theorem. It forces us to step away from our comfortable kinematic equations and think purely in terms of energy.

Constant Power vs. Constant Force

Imagine a particle of mass $m=0.2\text{ kg}$ sitting at rest. Suddenly, a force begins to act on it. But this isn't just any force; it's a force that delivers energy at a strictly constant rate. This rate is the power, $P=0.5\text{ W}$.

Here is where many students fall into a trap. They see a force causing motion and immediately reach for $v=u+at$. But wait! If the power $P=Fv$ is constant, and the velocity $v$ is increasing, the force $F$ must be decreasing. Since the force is changing, the acceleration is changing. Our trusty constant-acceleration kinematic equations are useless here. We need a more powerful tool.

Work-Energy Theorem

When forces are variable but we know about energy or power, the Work-Energy Theorem is our best friend. It states that the net work done on an object is exactly equal to its change in kinetic energy.

Since our particle starts from rest, its initial kinetic energy $K_i$ is zero. Therefore, the total work done on the particle simply becomes its final kinetic energy:

Now, how do we find the work done? We know the power $P$ is constant. Power is defined as the rate of doing work, $P=\frac{dW}{dt}$. If you are delivering $0.5\text{ Joules}$ of energy every single second, the total work done over a time $t$ is simply the power multiplied by the time.

This is the crucial logical bridge. We now have two different ways to express the same physical quantity—the work done. By equating them, we link the given parameters to the unknown velocity:

With our master equation established, the physics is complete, and only algebra remains. We want to find the final speed $v$. Let's isolate it:

Taking the square root of both sides gives us the explicit formula for velocity:

Now, we bring back the specific values from our problem: $P=0.5\text{ W}$, $t=5\text{ s}$, and $m=0.2\text{ kg}$. Substituting these into our formula:

The numerator simplifies beautifully. $2\times 0.5$ is exactly $1$, and $1\times 5$ is $5$.

Dividing by $0.2$ is the same as multiplying by $5$, so the term inside the square root becomes $25$.

The particle reaches a speed of $5\text{ m/s}$ after $5\text{ seconds}$. The elegance of the Work-Energy Theorem turns a potentially complex calculus problem into a straightforward algebraic calculation.