Question

b. Derive the formula for the kinetic
energy of an object of masS m,
of
moving with velocity v.​

Answer 1

We have to derive, formula of the kinetic energy! Let's see how to do!

Firstly we can say that kinetic energy is equal to the work done by an object.

I am saying that it because

Criteria 1st If a object hit from a point when at rest to another point slowly then of course it will go slowly & it have a displacement too

Criteria 2nd If a object hit from a point when at rest to another point with high work in fast motion then of course it will go fastly & it have a displacement too

• Wherever, be moved then of course it move with a kinetic energy.

(Remember, what is kinetic energy? Nice!)

Important: It is said that kinetic energy of an object moving with a certain velocity is equal to the work done on it to make it that acquire that velocity!

This state that kinetic energy is equal to the work done by an object.

__________________

Now let's derive!

⟶ Work done = Kinetic energy

(Symbols to denote the both)

⟶ W = ${\sf{E_k}}$

(Formula of work done)

⟶ F × S = ${\sf{E_k}}$

⟶ ${\sf{E_k}}$ = F × S

(Now finding value of F and S)

Finding the value of force!

According to the second law of motion, F is equal to product of mass & acceleration

Finding value of displacement!

As according to our criteria the object is at rest and have to displacement, initial velocity u & displacement s & final velocity v. Here time isn't given, then how to find displacement? Let's use equations of motion, according to third equation of motion, we find that

↪️ 2as = v² - u²

↪️ ${\sf{s \: = \dfrac{v^2 - u^2}{2a}}}$

Here, we know that u is 0, therefore

↪️ ${\sf{s \: = \dfrac{v^2 - 0}{2a}}}$

↪️ ${\sf{s \: = \dfrac{v^2}{2a}}}$

(Now let's carry on)

⟶ ${\sf{E_k}}$ = ma × ${\sf{ \dfrac{v^2}{2a}}}$

(Here a & a cancel each other, therefore)

⟶ ${\sf{E_k}}$ = ${\sf{m{\cancel{a}}}}$ × ${\sf{ \dfrac{v^2}{2{\cancel{a}}}}}$

⟶ ${\sf{E_k}}$ = ${\sf{\dfrac{1}{2}mv^2}}$

Henceforth, derived!