Comments on How can the kinetic energy equation be intuitively understood?
Parent
How can the kinetic energy equation be intuitively understood?
Momentum is proportional to an object's velocity, and kinetic energy is proportional to the square of its velocity $\dfrac{mv^2}{2}$. It's pretty intuitive that if object B is going twice as fast as object A and they have the same mass, B has twice the momentum. It's harder to grasp why it has four times as much kinetic energy. The math is clear enough, but for most of us it doesn't give a feel for why, for example, accelerating a car to 40 MPH gives it four times as much energy of motion as accelerating it to 20 MPH.
Looking at braking seems to help. If you're going twice as fast when you start to brake and your speed linearly goes to zero, it will take you twice as long at constant deceleration to decelerate to zero, and the average speed while braking will be twice as great (half your initial speed). This implies that the braking distance is proportional to the square of your initial speed, so you're applying a braking force for four times the distance. Similarly, accelerating a car to 40 MPH will take 4 times as much distance as accelerating to 20 MPH. But that doesn't quite get me to energy.
I think there's just one more step to making the relationship to kinetic energy obvious, but I don't have it.
Post
Your analysis seems pretty good. To take an equivalent but more traditional example, imagine we throw a ball upwards. Ignoring air resistance and approximating the gravitational force as constant, the ball will accelerate downward with constant acceleration. If the initial (upwards) velocity is $v_0$ and the downward acceleration is $a=-g$, the ball will reach the apex at time $t = v_0/a$ at which point it will have zero velocity and thus zero kinetic energy. It will have traveled a distance of $v_0^2/(2a)$. In general, work is force applied over a distance. In this case, the work done by the gravitational field is the (constant) force, $F = -mg$, applied over this distance, $v_0^2/(2a)$, giving a work done of $mv_0^2/2$.
Moving more towards your analysis, once we fix a constant force (and constant mass), then we know work is directly proportional to the distance, $d$, the mass travels, i.e. $W=Fd$. Your analysis shows that $d$ is quadratic in the initial velocity, thus work is quadratic in the initial velocity. I believe this was your goal. The slightly more detailed analysis in the previous paragraph will reproduce exactly the kinetic energy expression if you wanted more.
1 comment thread
0 comment threads