Question

explain the derivation of kinetic energy

Answer 1

Discussion
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) deri

Answer 2

Hy
Here is your answer....

The kinetic energy of an object is the energy an object has due to its motion. The formula for kinetic energy is:

KE=1/2 mv^2 , where m is the object’s mass and v is its velocity.

This formula is derived from the work-energy theorem, which states that the amount of work done on an object is equal to the change in its energy. Assuming potential energy is constant, Work is equal to the change in kinetic energy. This is expressed in the formula ∆KE=Fd. You can then use algebra or calculus to derive the kinetic energy equation.

Algebra Derivation:

∆KE=Fd
∆KE=mad (recall that F=ma)

Now, we use the kinematic equation: vf^2-vi^2= 2ad and solve for ad, then we substitute into the original equation.

∆KE=m((vf^2-vi^2)/2)

(vf is final velocity and vi is initial velocity)

We finally expand to get:

∆KE=1/2 mvf^2–1/2mvi^2

This is one of the countless ways to derive the kinetic energy theorem.

#Hope this helps u...