Bored and curious about kinetic energy

AI Thread Summary
The discussion centers on the kinetic energy formula, specifically questioning the origin of the 1/2 constant in Ek = 0.5mv^2. A test example involving a truck's mass, distance, and time led to a miscalculation of kinetic energy, prompting the inquiry. The 1/2 arises from the need to maintain consistency with the work-energy principle, particularly when considering the work done by a constant force. The relationship between acceleration, distance, and time is explained through basic physics concepts, including the equations of motion. The conversation concludes with an acknowledgment of the integration involved in deriving the kinetic energy formula.
mburt
Messages
52
Reaction score
0
So I was really bored today, and while pondering to myself I wondered why this isn't a valid formula for kinetic energy (say for a laterally moving truck):

E = mdx2 / t2

So I did a little test example, say a moving truck with:

m = 2000 kg
dx = 150 m
t = 10s

And plugged it in the formula and got 450 000 kg m2/s2 (J).

After remember the ACTUAL kinetic energy formula (Ek = 0.5mv^2), I was wondering where on Earth the 0.5 came from? In reality the truck has a kinetic energy of 225 000 J

So my question is simply this: (probably a basic one) Why is there a 1/2 constant in the kinetic energy formula?
 
Physics news on Phys.org
You need the 1/2 to be consistent with the law of conservation of energy.

Think about a falling object (ignoring air resistance).

The force acting on it is its weight = mg

The acceleration is constant = g

If it start from rest, you know that at time t

v = gt
x = 1/2 g t^2

The work done by its weight = force times distance = (mg) (1/2 g t^2)

The kinetic energy = 1/2 m (gt)^2

And those two quantities must be equal.
 
If it start from rest, you know that at time t

v = gt
x = 1/2 g t^2

Is this considered integration? It looks like you've worked backwards to find the original equation for the derivative v = gt. I've only done introductory calculus btw, and still learning.

And what would x represent here? EDIT: Never mind! I guess it means a distance, considering the units work out to meters! This makes sense too, considered F = mg, and Ek must be in J = Nm... so the distance aspect is (1/2 gt^2) and the "N" aspect = mg, so you must multiply them... awesome!

Thanks
 
Last edited:
Hi there, im studying nanoscience at the university in Basel. Today I looked at the topic of intertial and non-inertial reference frames and the existence of fictitious forces. I understand that you call forces real in physics if they appear in interplay. Meaning that a force is real when there is the "actio" partner to the "reactio" partner. If this condition is not satisfied the force is not real. I also understand that if you specifically look at non-inertial reference frames you can...
This has been discussed many times on PF, and will likely come up again, so the video might come handy. Previous threads: https://www.physicsforums.com/threads/is-a-treadmill-incline-just-a-marketing-gimmick.937725/ https://www.physicsforums.com/threads/work-done-running-on-an-inclined-treadmill.927825/ https://www.physicsforums.com/threads/how-do-we-calculate-the-energy-we-used-to-do-something.1052162/
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...
Back
Top