# 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 $d=\tfrac{1}{2} v t$. Feb 16 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. Feb 16 at 20:02

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.
• @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 $\bar{v} = \Delta x/t = \frac{1}{2} (v_f + v_i)$. Feb 16 at 14:41