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?
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$\begingroup$ Distance is the area under the triangle $v=a t$ and hence $d=\tfrac{1}{2} v t$. $\endgroup$– John AlexiouFeb 16 at 15:04
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$\begingroup$ I'm not sure why this question was closed as homework-like. It definitely seems to be asking about a conceptual difficulty. $\endgroup$– Michael SeifertFeb 16 at 20:02
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1 Answer
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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.
answered Feb 16 at 13:45
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$\begingroup$ 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? $\endgroup$ Feb 16 at 14:12
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1$\begingroup$ @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 $\bar{v} = \Delta x/t = \frac{1}{2} (v_f + v_i)$. $\endgroup$ Feb 16 at 14:41