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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.

Qmechanic
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Pancake_Senpai
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4 Answers4

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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!

Steven
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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 $KE=mv^2/2$

J.G.
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Alex S
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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!

Daniel Lee
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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.

J.G.
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