FFR - Kinetic Energy

Calculus Students:

If you are really curious, and want to know why the constant in the equation is $\frac{1}{2}$, it’s because of the calculus involved.

Just like we can write the work equation in integral form, we can write the kinetic energy equation in integral form:

ΔK = ∫ from v₁ to v₂ mv dv = ½m(v₂²−v₁²), for one-dimensional motion of constant mass.

For an initially resting particle, K(v)=∫₀ᵛmu du=½mv².

If you don’t know calculus, here’s a quick definition:

${\int }_a^{b}C{x}^{n}dx=C{\int }_a^b{x}^{n}dx$

$C{\int }_a^b{x}^{n}dx=C{\frac{x^{n+1}}{n+1}|}_a^b$ $C{\frac{x^{n+1}}{n+1}|}_a^b=C\left(\frac{b^{n+1}}{n+1}-\frac{a^{n+1}}{n+1}\right)$

That’s the general form for integrating a basic function. A bit complicated, I know, but if we can assume that we start the integral at 0, we can simplify it a bit:

${\int }_0^{a}C{x}^{n}dx=C\left(\frac{a^{n+1}}{n+1}\right)$ (Equation 4.4.5)​

There we go, a nice, simple way of solving a basic integral.

Now how does this relate to kinetic energy? Take a look at our integral solution for $KE$.

ΔK = ∫ from v₁ to v₂ mv dv = ½m(v₂²−v₁²), for one-dimensional motion of constant mass.

Here the integration variable is speed/velocity component u, the power is 1, and the constant coefficient is mass m.

K(v)=m∫₀ᵛu du=m[v²/2−0]=½mv².

Re-writing that, we have:

$KE=\frac{1}{2}m{\left(\overrightarrow{v}\right)}^2$

Step 1: Define both terms individually and simplify

To determine this ratio, we must first transform both variables into forms with as many parts in common. In this case, momentum and velocity will be our desired variables. For kinetic energy we use the kinetic energy definition in tandem with the momentum definition:

$\mathrm{\Delta }KE=\frac{1}{2}{m}_2{\left({\overrightarrow{v}}_2\right)}^2-\frac{1}{2}{m}_1{\left({\overrightarrow{v}}_1\right)}^2$

As mass is constant, we can say:

$\mathrm{\Delta }KE=\frac{1}{2}m\left[{\left({\overrightarrow{v}}_2\right)}^2-{\left({\overrightarrow{v}}_1\right)}^2\right]$

For momentum, we can simply use the definition of momentum:

$\mathrm{\Delta }\overrightarrow{p}=m{\overrightarrow{v}}_2-m{\overrightarrow{v}}_1$

Again, mass is constant:

$\mathrm{\Delta }\overrightarrow{p}=m\left({\overrightarrow{v}}_2-{\overrightarrow{v}}_1\right)$

Step 2: Form the ratio

Now we take our new definitions and place them atop one another:

$\frac{\mathrm{\Delta }KE}{\mathrm{\Delta }\overrightarrow{p}}=\frac{\frac{1}{2}m\left[{\left({\overrightarrow{v}}_2\right)}^2-{\left({\overrightarrow{v}}_1\right)}^2\right]}{m\left({\overrightarrow{v}}_2-{\overrightarrow{v}}_1\right)}=\frac{1}{2}\frac{{\left({\overrightarrow{v}}_2\right)}^2-{\left({\overrightarrow{v}}_1\right)}^2}{{\overrightarrow{v}}_2-{\overrightarrow{v}}_1}$

As the numerator is a difference of squares, we can expand that into:

$\frac{\mathrm{\Delta }KE}{\mathrm{\Delta }\overrightarrow{p}}=\frac{1}{2}\frac{\left({\overrightarrow{v}}_2+{\overrightarrow{v}}_1\right)\left({\overrightarrow{v}}_2-{\overrightarrow{v}}_1\right)}{{\overrightarrow{v}}_2-{\overrightarrow{v}}_1}$

This allows the simplification:

For signed one-dimensional components, ΔK/Δp = ½(v₂+v₁), provided v₂ ≠ v₁.

This is noteworthy as it is the average of the two velocities, a rather linear relationship.

Step 1: Investigate the kinetic energy question to see what is needed

The kinetic energy increase can be given by:

ΔK = ∫ from v₁ to v₂ mv dv = ½m(v₂²−v₁²), for one-dimensional motion of constant mass.

As we know mass and we are given a position function, we can find the velocity function by differentiating the position function:

$\overrightarrow{v}=\frac{dx}{dt}$​
$\overrightarrow{v}=\frac{d}{dt}3t^2+t$​
$\overrightarrow{v}=6t+1$

Now we have a definition of velocity as a function of time. This allows us to solve for Kinetic Energy as a function of time, instead of velocity.

Step 2: Plug the velocity equation into the integral and solve

ΔK = ∫ from v₁ to v₂ mv dv = ½m(v₂²−v₁²), for one-dimensional motion of constant mass.

Using v(t)=6t+1 m/s and a=6 m/s², ΔK=∫ from t=1 s to t=2 s m v(t)a dt.

Now we need to figure out three things, v_1, v₂, and dv. We find the v’s through the velocity equation, v₁ is 7m/s and v₂ is 13m/s. We can find dv from the acceleration equation:

$\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}=\frac{d}{dt}6t+1=6$​
$d\overrightarrow{v}=\overrightarrow{a}dt$

$\overrightarrow{a}$ is therefore $6\text{m}/{\text{s}}^2$.

Now we can solve the integral.

Plugging in all we now know, the integral becomes:

ΔK = 60∫₁²(6t+1) dt.

We can evaluate this integral as:

60∫₁²(6t+1)dt = 60[3t²+t]₁².

60[(3·2²+2)−(3·1²+1)] = 60(14−4).

Therefore, Kinetic energy is:

ΔK = 600 J = ½(10 kg)[(13 m/s)²−(7 m/s)²].

Increase in kinetic energy = 0.600 kJ.

Section breakdown

1. Kinetic Energy
2. Calculus Galore
3. Example 4.4.1 - Abstract Fun with Imaginary Objects
4. Example 4.4.2 - A Cool Calculus Problem
5. Chapter Equations
6. Chapter Problems