Kinetic Energy and Conservation of Energy, Part 1

I’ve been wanting to systematically go through a physics text, refreshing my knowledge and filling in gaps. I found The Feynman Lectures on Physics, freely available online, based on lectures from the great physicist and physics popularizer, Richard Feynman. As I read Chapter 4: “Conservation of Energy,” I wonder if I can derive the formula for kinetic energy, the energy an object has due to its motion. My primary motivation is that while it is relatively easy for me to grasp why the gravitational potential energy is U=mgh (that is, weight times height), I have a much harder time intuitively understanding why the kinetic energy should be proportional to the velocity squared: Kmv².

Problem: Given the formula for gravitational potential energy, can I use the principle of conservation of energy (of a projectile) to derive the equation for its kinetic energy?

The total energy of an object should be the sum of its potential energy and kinetic energy: EUK, and it should remain constant.

\displaystyle E_1 = E_2

\displaystyle U_1 + K_1 = U_2 + K_2

If we pick the right times, we can make this much simpler. If an object is launched vertically (from, say, height = 0), it initially has kinetic energy but no potential energy (U1 = 0). At the peak of its trajectory, it will be motionless for an instant — it will have potential energy but no kinetic energy (K2 = 0). Therefore,

\displaystyle 0 + K_1 = U_2 + 0

\displaystyle K_1 = U_2

In a constant gravitational field, the potential energy is U=mgy, where m is the mass, g is the acceleration due to gravity, and y is the height. At the second time point,

\displaystyle U_2 = mgy_2

\displaystyle U_2 = K_1 = mgy_2

where y2 is the maximum height it will reach. We can find this from the equations of projectile motion, given constant acceleration a:

\displaystyle y = \frac{1}{2}at^2 + v_1t + y_1

The acceleration a is the acceleration due to gravity, oriented down:

\displaystyle y = -\frac{1}{2}gt^2 + v_1t + y_1

And the initial height was zero:

\displaystyle y = -\frac{1}{2}gt^2 + v_1t 

The velocity is the derivative of position (with respect to time):

\displaystyle v = y' = -\frac{1}{2}\cdot 2gt + v_1

\displaystyle v  = -gt + v_1

At the peak, the velocity and therefore kinetic energy will be zero.

\displaystyle 0 = -gt_2 + v_1

\displaystyle gt_2 = v_1

\displaystyle t_2 = \frac{v_1}{g}

As an aside, this should have been intuitively obvious: acceleration is change in velocity over time, so time should be change in velocity over acceleration.

\displaystyle  a = \frac{dv}{dt}

\displaystyle  \bar{a} = \frac{\Delta v}{\Delta t}

If the acceleration is constant, then a = ā (the instantaneous acceleration equals the mean acceleration).

\displaystyle  a = \frac{\Delta v}{\Delta t}

\displaystyle \Delta t = \frac{\Delta v}{a}

\displaystyle t_2-0 = \frac{0-v_1}{-g}

\displaystyle t_2 = \frac{v_1}{g}

Let’s substitute this back into the projectile equation.

\displaystyle y_2 = -\frac{1}{2}g{t_2}^2 + v_1t

\displaystyle y_2 = -\frac{1}{2}g\left({\frac{v_1}{g}}\right)^2 + v_1\left({\frac{v_1}{g}}\right)

\displaystyle y_2 = -\frac{1}{2}g\cdot\frac{{v_1}^2}{g^2} + v_1\left({\frac{v_1}{g}}\right)

\displaystyle y_2 = -\frac{1}{2}\frac{{v_1}^2}{g} + v_1\left({\frac{v_1}{g}}\right)

\displaystyle y_2 = -\frac{1}{2}\frac{{v_1}^2}{g} + \frac{{v_1}^2}{g}

\displaystyle y_2 = \frac{{v_1}^2}{g}\left({-\frac{1}{2} + 1}\right)

\displaystyle y_2 = \frac{{v_1}^2}{2g}

We can now substitute this back into our equation for K1:

\displaystyle K_1 = mgy_2

\displaystyle K_1= mg\frac{{v_1}^2}{2g}

\displaystyle K_1 = m\frac{{v_1}^2}{2}

 \displaystyle K_1 = \frac{1}{2}m{v_1}^2 \ \blacksquare 

This is indeed the formula for (nonrelativistic) kinetic energy.

The calculation was straightforward, but unfortunately, it didn’t really give me an intuitive understanding of the kinetic energy equation. I have some ideas, which I’ll explore in another post.

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