# Does $Power=F\cdot V$ imply $K.E=mv^2$?

The work done on a body with mass is equal to the constant force applied on the mass, multiplied by the distance over which that force is applied ($$W = Fd$$). Dividing both sides by the time over which the force is applied, $$t$$, we get $$P = Fv$$ where $$P$$ is the power (rate of work done) and $$v$$ is the average speed of the body. However, $$F$$ is equal to the final momentum of the body divided by the time it takes to reach it. So $$F = mv/t$$. Therefore, $$P=\frac{mv^2}{t}$$. Finally multiplying both sides by $$t$$: $$W = mv^2$$, where $$W$$ is supposed to equal $$\frac{mv^2}{2}$$. Why is this incorrect?

• Distance is the area under the triangle $v=a t$ and hence $d=\tfrac{1}{2} v t$. Commented Feb 16, 2023 at 15:04
• I'm not sure why this question was closed as homework-like. It definitely seems to be asking about a conceptual difficulty. Commented Feb 16, 2023 at 20:02

You're conflating instantaneous velocity (and power) with final velocity (and power).

If we have a constant force being applied to an object starting at rest, then the instantaneous power being applied is $$P = F v$$. That means that the power is small at the beginning of the motion (when $$v$$ is small) and increases up to $$P = F v_f$$ by the end of the motion (where $$v_f$$ is the final velocity.) It is not the case that the power being applied is equal to $$P = F v_f$$ throughout the motion, and so the final energy should not be expected to be $$F v_f t = (m v_f/t) v_f t = m v_f^2$$.

Instead, we can note that the average power over the motion is equal to the force multiplied by the average velocity over the motion: $$\bar{P} = F \bar{v}$$. Moreover, the average velocity over the motion is one-half of the final velocity: $$\bar{v} = \frac12 v_f$$. The total power imparted will then be $$\bar{P} t = \frac12 m v_f^2$$, as expected.

• Thank you for the answer. I understand what you mean with the final power not being the power for the entire motion. However, why is the average velocity over the motion half of the final velocity? Commented Feb 16, 2023 at 14:12
• @YohannesTimket: It's a general feature of uniformly accelerated motion: the average velocity over the motion is the average of the initial and final velocity. In your case, the initial velocity is zero and so $\bar{v} = \frac12 v_f$. To see the general result, note from the kinematic equations that $\Delta x = v_i t + \frac{1}{2} at^2 = v_i t + \frac{1}{2} (v_f - v_i) t = \frac12 (v_f + v_i) t$. So $\bar{v} = \Delta x/t = \frac{1}{2} (v_f + v_i)$. Commented Feb 16, 2023 at 14:41