Question

Derive the relationship- [2] Kinetic Energy = ½ mv2 Where m = mass of the body And v = velocity of the body​

Answer 1

Answer:

K.E= 1/2mv2

Explanation:

W=F×s --------(1)

from equation of motion

v2-u2=2as

s= (v2-u2)/2a ---------(2)

F= ma --------(3)

substituting (2)&(3) in (1), we get

W= ma×(v2-u2)/2a

W = 1/2m(v2-u2)

(u=0) object start at rest so initial velocity is zero

W = 1/2mv2

hence proof.

Answer 2

⇝ To Derive :

$\text{K.E.}= \dfrac{1}{2} \text{mv}^{2}$

Where,

• m = Mass of the Body

• v = Velocity of the Body

----------------------

⇝ Derivation :

★ Using Newton's First Law :

• F = ma - - - - (1)

★ Work Done by Force is given by :

• W = F.ds - - - - (2)

✏ Putting (2) in (1) :

$:\longmapsto\text{W = ma.ds}$

$\bf:\longmapsto W = ma \times s \: - - - - (3)$

★ Using v² - u² = 2as :

• $\bf s = \frac{ {v}^{2} - {u}^{2} }{2a}$ - - - - (4)

✏ Putting (4) in (3) :

$:\longmapsto\text W = \text{m} \cancel{\text a} \times \dfrac{ {\text v}^{2} - {\text u}^{2} }{2 \: \cancel\text a}$

$:\longmapsto\text W = \dfrac{1}{2} \text m( {\text u}^{2} - {\text v}^{2})$

✏ When initial velocity = u = 0

$:\longmapsto\text W = \dfrac{1}{2}\text {mv} {}^{2}$

★ This work is stored as kinetic energy :

Hence,

$\Large\underline{\pink{\underline{\frak{\pmb{K.E.= \dfrac{1}{2}m \bf {v}^{2} }}}}}$