Quote:
Originally Posted by lastcardcharlie
Help a physics idiot out. If kinetic energy equals mv^2/2, then it takes more energy for something to accelerate from 2 mph to 3 mph than it does for 1 mph to 2 mph, which seems counter-intuitive. Am I missing something?
That is because to go from 2 to 3 under say constant acceleration you take more distance that the force accelerating it is applied, since vf^2=vi^2+2*a*x (F=m*a , a acceleration)
So the work that is defined as W=F*x is more than from 1 to 2. I can "prove"/support why work is defined like that because i can argue that the "effort" to move something under constant speed against friction depends proportionally on for how much time you try to move it at constant speed (hence depends on distance proportionally) and how much of it you move, for clearly 2 times something done in parallel takes twice the effort and stacking them up to each other creating the need for double the force (friction is proportional to normal force or weight in this case) gets the same result (at constant speed we have force = force of friction opposing the effort say). So work is proportional to distance and force. That it is precisely F*x is arbitrary but we can see how this connects in experiments to the change of the quantity 1/2*m*v^2 which connects the effort with the result ie the kinetic energy change naturally defined as the work of the applied force. Also on the reverse the capacity of an object to do work because of its motion and mass is defined as kinetic energy. Eg a moving object pressing a spring as it decelerates and producing the work needed to compress it results in potential energy of the spring (ie the capacity of spring to do work now if released) the exactly same result as if one applied a force to it slowly to compress it. Experiments show that work is done on all these processes and it is the same leading to the concept of conservation of energy as it transforms from one kind to another through work.
Basically it comes from F=m*a= m*dv/dt. dW=F*dx=m*dv/dt*dx=m*dv/dt*dx/dt*dt==m*v*dv/dt*dt=1/2*m*d(v^2)/dt *dt
so dw/dt=1/2*m*d(v^2)/dt=d(1/2*m*v^2)/dt
=d(Ekinetic)/dt
or rate of work ie power is rate of change of kinetic energy.
(work is defined as dW=F*x or better the element of work is the dot product of force times differential displacement)