prove that kinetic energy of a body moving with speed v is equal to 1/2mv²
Answers
Answer:
The answers which I have seen below are absolutely true! Using Newton’s relationship between velocities, acceleration, and displacement, however, is a traditional off-the-book method. I recently discovered a different proof that relies on infinitesimal calculus.
When a real power (any distribution dependent on t), P, is exerted on a body for a fixed time interval, t, the total energy imparted to the body is:
E = Pt
Let’s assume that the energy imparted to the body is purely translational, non-rotational (therefore non-inertial) and is not being lost to the environment.
Therefore, the Energy distribution will be the definite riemann summation (the definite integral) of the single-variable Power function, bounded from [0, t]
So, Energy = ⨛P dt
Power is defined as E/t
P = E/t, where E is the energy transferred to the body or the work done on the body, which is equal to Force x displacement
P = Fd/t. The d/t factor can be substituted by v, the average velocity of the body during the motion (note that this is applicable throughout the motion, even if the velocity is instantaneous, which is true in this case)
Therefore, P = Fv
E = ⨛P dt = ⨛Fv dt = ⨛ma (dx/dt) dt = ⨛ma dx
Here, I essentially substituted P = Fv, opened up F to ma, and v to (dx/dt). When I did this, (dt * dx)/dt canceled out to leave dx
E = ⨛ma dx
a = dv/dt, v = dx/dt. Making dt the subject of the equation, dt = dv / a = dx / v. You will get a = v * dv/dx
Plug this back into the integral:
E = ⨛ma dx = ⨛m v (dv/dx) dx
dx*dv/dx cancel to leave dv
The Integral then boils down to:
E = ⨛mv dv
m, the mass, is not dependent on v, and hence can be factored out of the integral.
E = m⨛v dv = ½ mv² +k
K is 0 because we don’t expect a body at rest to possess any kinetic energy, since Kinetic energy is primarily a measure of speed.
Therefore, kinetic energy = ½ mv²