drevie an equation of kinetic engery
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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 − v022
Combine the two expressions.
ΔK = m ⎛
⎝v2 − v02⎞
⎠2
And now something a bit unusual. Expand.
ΔK = 1 mv2 − 1 mv0222
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 · drdt
Rearrange the differential terms to get the integral and the function into agreement.
ΔK = m⌠
⌡dv · drdtΔK = m⌠
⌡dr · dv dtΔK = 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 mv0222
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
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 − v022
Combine the two expressions.
ΔK = m ⎛
⎝v2 − v02⎞
⎠2
And now something a bit unusual. Expand.
ΔK = 1 mv2 − 1 mv0222
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 · drdt
Rearrange the differential terms to get the integral and the function into agreement.
ΔK = m⌠
⌡dv · drdtΔK = m⌠
⌡dr · dv dtΔK = 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 mv0222
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
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PE=mgh this is the answer
jabroonmathew:
12h
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