proof kinetic energy in equation
Answers
Kinetic energy is a simple concept with a simple equation that is simple to derive. Let's do it twice.
Derivation using algebra alone (and assuming acceleration is constant). Start from the work-energy theorem, then add in Newton's second law of motion.
ΔK = W = FΔs = maΔs
Take the the appropriate equation from kinematics and rearrange it a bit.
v2 = v02 + 2aΔs
aΔs = v2 − v02
2
Combine the two expressions.
ΔK = m ⎛
⎝ v2 − v02 ⎞
⎠
2
And now something a bit unusual. Expand.
ΔK = 1 mv2 − 1 mv02
2 2
If kinetic energy is the energy of motion then, naturally, the kinetic energy of an object at rest should be zero. Therefore, we don't need the second term and an object's kinetic energy is just…
K = ½mv2
Derivation using calculus (but now we don't need to assume anything about the acceleration). Again, start from the work-energy theorem and add in Newton's second law of motion (the calculus version).
ΔK = W
ΔK = ⌠
⌡ F(r) · dr
ΔK = ⌠
⌡ ma · dr
ΔK = m ⌠
⌡ dv · dr
dt
Rearrange the differential terms to get the integral and the function into agreement.
ΔK = m ⌠
⌡ dv · dr
dt
ΔK = m ⌠
⌡ dr · dv
dt
ΔK = m ⌠
⌡ v · dv
The integral of which is quite simple to evaluate over the limits initial speed (v) to final speed (v0).
ΔK = 1 mv2 − 1 mv02
2 2
Naturally, the kinetic energy of an object at rest should be zero. Thus an object's kinetic energy is defined mathematically by the following equation…
K = ½mv2
Thomas Young (1773–1829) derived a similar formula in 1807, although he neglected to add the ½ to the front and he didn't use the words mass and weight with the same precision we do nowadays. He was also the first to use the word energy with its current meaning in a lecture on collisions given before the Royal Institution.