Speed of Particle Accelerated by Constant Power

To determine the speed of the particle after a duration of $5\text{ s}$, we apply the work-energy theorem.

The power delivered by the force to the particle is constant and given by $P = 0.5\text{ W}$.

The total work done $W$ on the particle in time $t$ by this constant power is: $W = \int_0^t P \, dt = P \cdot t$

According to the work-energy theorem, the net work done on the particle is equal to the change in its kinetic energy: $W = \Delta K = K_f - K_i$

Since the particle starts from rest, its initial velocity is $u = 0\text{ m s}^{-1}$, which means: $K_i = 0$

Let $v$ be the final speed of the particle at time $t$. The final kinetic energy is: $K_f = \frac{1}{2}mv^2$

Equating the work done to the final kinetic energy: $P \cdot t = \frac{1}{2}mv^2$

Rearranging the equation to solve for $v$: $v = \sqrt{\frac{2Pt}{m}}$

Substitute the given values:

• $m = 0.2\text{ kg}$
• $P = 0.5\text{ W}$
• $t = 5\text{ s}$

$v = \sqrt{\frac{2 \times 0.5 \times 5}{0.2}} = \sqrt{\frac{5}{0.2}} = \sqrt{25} = 5\text{ m s}^{-1}$

Thus, the speed of the particle after $5\text{ s}$ is $5\text{ m s}^{-1}$.