Question

Dirive kinetic energy.

Answer 1

ANSWER:-
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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Answer 2

$\mathfrak{Derivation \: of \:Kinetic \: Energy}$

$\mathsf{Things \: to \: know \: before\: Derivation}$

$\mathsf{\implies Work done = Fs}$

$\mathsf{\implies v² = u² + 2as}$

$\mathsf{\implies s = \frac{v² - u²}{2a}}$

$\mathsf{\implies u = 0 m/s \: for \:a \: body \: starting \: from \: rest}$

$\mathsf{\implies Work \: Done = \: Energy}$

$\mathsf{\implies Kinetic \: Energy \: is \: also \: written \: as \: E_k}$

$\mathsf{\implies Force = mass \: × \: acceleration \: = ma}$

$\mathsf{Derivation}$

$\mathsf{\hookrightarrow E_k = Work done = Fs }$

$\mathsf{\hookrightarrow \: = \: Fs \: = ma × s }$

$\mathsf{\hookrightarrow E_k = m × \frac{v² - u²}{2a} × a}$

$\mathsf{\hookrightarrow E_k = \frac{1}{2}mv²}$