Question

kinetic energy derivation

Answer 1

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

Answer 2

Answer:

Explanation:consider a body of mass m initially at rest

, `, [v^2]-[u^2]=2as

As u=0(initially at rest)

,`, v^2=2as

Dividing both sides by 1/2*m, we get

1/2*mv^2=2as*(1/2)m

Therefore 1/2mv^2=fs=w

(F=ma and fs=w)