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To derive the equation I used a system where a particle starts at rest, and then has a constant force applied to it so that it accelerates with a constant acceleration. At time $t=T$ the particle has velocity $v$, acceleration $a$, kinetic energy $E_k$, mass $m$ and a force $F$ being applied on it.

$$F = ma$$ $$F = m \frac{\Delta v}{\Delta t} = \frac{m(v - 0)}{\Delta t} = \frac{mv}{\Delta t}$$ $$F \Delta t = mv$$

Work must be done on the particle for it to have kinetic energy.

$$\Delta E_k = F \Delta d$$ $$F = \frac{\Delta E_k}{\Delta d}$$

Combining the two equations:

$$\frac{\Delta E_k \ \Delta t}{\Delta d} = mv$$ $$\frac{\Delta E_k}{v} = mv$$ $$E_k - 0 = mv^2$$ $$E_k = mv^2$$

I appear to have a missing $\frac{1}{2}$ in my derivation. Could anyone please point me in the right direction or show me any mistake I've made? I appreciate the complete derivations for this are all online but I'd rather not just give myself the answer.

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  • $\begingroup$ I believe you need to go through calculus to arrive at the factory if 1/2, does it need to be through algebraic manipulations? $\endgroup$ – QtizedQ Jul 12 '17 at 11:25
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    $\begingroup$ Related, if not a dupe of, physics.stackexchange.com/q/27847 (and see links therein) $\endgroup$ – Kyle Kanos Jul 12 '17 at 11:45
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You can derive this without any calculus methods. From the Galileo's equation we have: $v_f^2=v_i^2+2ad$, where $v_f$ is the final velocity, $v_i$ the initial velocity, a the acceleration produced by a constant force and d the distance. Multiply the equation with $m/2$ and we got $mv_f^2/2-mv_i^2/2=2mad$. But $2mad$ is the work $W$. So $mv_f^2/2-mv_i^2/2=W$. Also from the fundamental work-energy theorem we have $Ef-Ei=W$, where Ef,Ei are the energies corresponding to the final,initial states. So in accordance to that, we deduce that $KE=mv^2/2$

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  • $\begingroup$ In regards to both your answer and the other answer by @Steven , I understand your answers but I don't understand what I did wrong in my derivation. I must have made an error otherwise I would have had my missing $\frac{1}{2}$. $\endgroup$ – Pancake_Senpai Jul 12 '17 at 11:46
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    $\begingroup$ Then you made a mistake when you said that $\Delta t/\Delta d=1/v$. This assumption would have been correct if the mass would have been going with constant velocity, but is not the case here $\endgroup$ – Alex S Jul 12 '17 at 11:53
  • $\begingroup$ $\frac{\Delta t}{\Delta d} = \frac{1}{\Delta v}$, but when the object starts from rest then $\Delta v = $ final velocity of particle, which I'm representing by $v$. $\endgroup$ – Pancake_Senpai Jul 12 '17 at 12:11
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    $\begingroup$ This equation of motion is not correct. When a mass goes with a constant acceleration, the correct equation of motion is $\Delta d=va*\Delta t$, where $\Delta d$ is the difference $xf-xi$ corresponding with the final and initial coordinates of the mass, va is the average velocity $va=(vi+vf)/2$ and $\Delta t$ is the time. In your case, $va$ is not $vf$ is $vf/2$ or according to your notation is $v/2$ and you got the correct solution with $2/v$ $\endgroup$ – Alex S Jul 12 '17 at 12:20
  • $\begingroup$ It is a big difference between the equation of motion of a mass going with constant velocity and the equation of motion of a mass going with constant acceleration. The first equation is $\Delta d=v*\Delta t$ and the last one is $\Delta d=va*\Delta t$. Again, you applied the first one, not the second. You perhaps don't realize that because $vi=0$ $\endgroup$ – Alex S Jul 12 '17 at 12:26
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One way I see is to take the limit of infinitesimally small $\Delta d$, $\Delta t$, and $\Delta E_k$. Then, resuming from this step:

$\Delta E_k \frac{\Delta t}{\Delta d} = mv$

taking the limit of small $t$, $d$, and $\Delta E_k$, re-write them as differential units, and use $ dd / d t = d v$ (forgive the awkwardness of the differential unit of distance being $dd$):

$ \frac{ d E_k}{d v} = mv $

And then,

$ d E_k = mv \ d v$

Integrate both sides,

$\int d E_k = \int m v d v = \frac{mv^2}{2}= E_k$

As we had hoped to show!

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  • $\begingroup$ As I commented above, after looking at your answer I understand the calculus approach, but when the object starts from rest is it applicable? It's started from rest, so $\Delta v = v$ [final velocity]. The differential can be eliminated and we can move on without it. $\endgroup$ – Pancake_Senpai Jul 12 '17 at 12:13
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    $\begingroup$ @Pancake_Senpai But $\frac{\Delta d}{\Delta t}$ is not equal to $v$. It is the average velocity, which for constant acceleration is $\frac{v_f+v_i}{2}=\frac{v+0}{2}$. There is your 1/2 factor. $\endgroup$ – Bill N Jul 12 '17 at 14:23
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I'll identify your error: Your formula for $F$ assumes constant acceleration from initial speed $0$ to final speed $v$. Then the mean speed is $v/2$, so $\Delta d=v\Delta t/2$. If you carry on from there, you'll get the right result.

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You are correct until you got to`

$$\frac{\Delta E_k \ \Delta t}{\Delta d} = mv$$

Remember that $\frac{\Delta d}{\Delta t} = $ Average Velocity (And I'll explain why further down)

When we consider constant acceleration, average velocity is denoted as $\frac{V_o+V_f}{2}$

In your case, the particle is at rest. So initial velocity is zero.

We can rewrite work as

$$\frac{\Delta E_k}{{v_f}/{2}} = mv_f$$ $$E_k - 0 = \frac{1}{2}mv_f^2$$ So the Kinetic Energy equation will be $$E_k= \frac{1}{2}mv_f^2$$

Of course, there are many other ways to derive the formula.
One easy way to derive it is integrating the force in terms of distance, but I don't believe your problem requires calculus to obtain the formula.
Another way we could solve it is using Galileo's equation, but honestly, that makes it more confusing if one doesn't know where to derive that equation from (like me).

If you think about it regarding Work, you know that it is
$$Work = Force \cdot Distance$$

The reason why we know that $E_p = mgh$ is that we know what the initial position is. We know how much energy will be exerted from that height if it falls down to the ground. But it wouldn't work the same way if you wanted to find kinetic energy from the particle's current position. Instead, some problems will tell you to find the energy given time. You need its instantaneous velocity. We can use calculus to find instantaneous velocity, but we know one thing that allows us to find it without calculus: acceleration is constant.

This means that Average Velocity IS Instantaneous velocity. And it just so happens that the particle starts at rest. Does that mean the final velocity equals the instantaneous velocity? No. If there were a problem that asks you to find the kinetic energy while it was in motion, this would be this full equation:

$$E_k=m\cdot(V_f - V_o)\cdot(\frac{V_o+V_f}{2})$$

where $V_f - V_o$ is the change in velocity and $\frac{V_o+V_f}{2}$ is the average velocity

It was just convenient to have an application where the particle is at rest and acceleration is constant.

Hopefully, this explanation helps!

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  • $\begingroup$ That is not the correct expression for average velocity. $\endgroup$ – Aaron Stevens Jun 2 '18 at 5:21
  • $\begingroup$ @AaronStevens I'm sorry if my notations are weird, but it shouldn't impede any understanding at all. And I'm just a high school student. I was taught using that equation. So can you please take back the downvote? $\endgroup$ – Daniel Lee Jun 2 '18 at 6:53
  • $\begingroup$ @AaronStevens I think it's OK that he used the expression $v_{avg}=\frac{v_{0}+v_{f}}{2}=\frac{\Delta d}{\Delta t}$, because the question specifies constant acceleration, and though there's unconventional yet intuitive $d$ for displacement (usual is $s$), it's standard high-school algebraic physics which is based on simplifications appropriate for the question. But I have no idea what he means by "$\frac{V_o+V_f}{2}$ is instantaneous velocity", because that's wrong. Also there's some usage of the 'star product', which isn't acceptable in physics, but isn't a major problem. $\endgroup$ – user191954 Jun 2 '18 at 9:12
  • $\begingroup$ @Chair Oops yes I meant to say instantaneous velocity! $\endgroup$ – Aaron Stevens Jun 2 '18 at 13:26
  • $\begingroup$ @AaronStevens technically, that isn't the exact expression for average velocity either; the answer should have specified constant acceleration to prevent a confusion, and after that, I forgot to mention that in one place when I edited it. $\endgroup$ – user191954 Jun 2 '18 at 17:50

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