Okay, I've always been someone who doesn't learn anything unless I can prove it. In this case, I set out to prove the Kinetic Energy formula and (wrongly?) found that it was based on constant acceleration. (I didn't use calculus, I proved the idea of calculus as I went, using geometry and basic algebra) I'm really not entirely sure how to ask this question, but I'll try to explain as much as I can.
So a weight is connected to a spring, you pull the weight down $x$ distance, and you have to find the acceleration at that point along with the equation of motion. (This is for a controls class revolving around oscillations/frequency/settling time etc.) Now I didn't think this would be a problem, but since the answer involves equating kinetic energy of the spring and the system as well as equivalent masses and spring constants, I couldn't wrap my head around it. If Kinetic energy is based on constant acceleration and the spring force (and therefore, acceleration) is constantly changing, how can you equate it so simply?
Thanks, and sorry if I made some problematic assumptions or there was a complete lack of clarity.
Best Answer
These equations use calculus but give a succinct proof that the force is always the change in kinetic energy with respect to distance, whether or not the acceleration in constant:
$$ F = ma = m \frac{dv}{dt} = mv \frac{dv}{dt} \frac{1}{v} = mv \frac{dv}{dt} \frac{1}{dx/dt} = mv \frac{dv}{dt} \frac{dt}{dx} = mv \frac{dv}{dx} = \frac{d}{dx} \left( \frac{1}{2} m v^2 \right) = \frac{d(KE)}{dx}.$$
Integrating with respect to $x$ gives $$ \int_{x_1}^{x_2} F(x)\, dx = KE(x_2) - KE(x_1) = \Delta KE.$$
If the force is conservative then the LHS is the potential energy difference $U(x_1) - U(x_2)$, and rearranging gives
$$ KE(x_1) + U(x_1) = KE(x_2) + U(x_2),$$ which is the statement of conservation of energy.