Question

The speed of a moving body is halved. What is the change in its K.E.?

Answer 1

Here you have to consider the speed as velocity,

We know that ,

K.E (Kinetic Energy) = 1/2 * m * v² = 1/2mv²,

Now, Let at the beginning of the time velocity be, v,

Then K.E (1) = 1/2mv²,

Given that velocity has become halved, => new velocity = v/2,

Substituting this in the formula again, Now new K.E is K.E(2) = 1/2m*(v²/4) = 1/8mv²,

Now to calculate change in K.E, Subtract K.E(1) from K.E(2),

=> ∆K.E or change in K.E = 1/8mv² - 1/2mv²,
=>∆K.E = -3/8 mv²,
=> ∆K.E = -3/4 * K.E(1),

So therefore change in K.E equals -3/8mv² or -3/4 times the initial K.E,

Hope you understand,

Have a great day,

Thanking you,

Bunti 360 !

Answer 2

Heya User,

--> Kinetic Energy = 1/2 [ mass ] * [ velocity ]² = 1/2 mv²

--> The velocity is considered to be the one with which it is travelling|| the final velocity........

--> Let us represent --> u = Initial Velocity ; v = Final Velocity..

Now,
----> v = u/2 ;
 
Applying K.E. formula -->
---> Initial K.E. = 1/2 m.u²
---> Final K.E.  = 1/2 m.[ u/2 ]²

.'. --> Change in K.E. = $\frac{Final\:K.E.\:-\:Initial\:K.E.}{Initial\:K.E.} * 100$
==> Change in K.E. = $\frac{ \frac{1}{2}m \frac{u^{2} }{4}- \frac{1}{2} m u^{2} }{ \frac{1}{2} m u^{2} } * 100$
==> Let 1/2 mu² = k;

Then, Change in K.E. = $\frac{ \frac{k}{4} -k}{k} * 100 = -3k/4k * 100$

--> Change in K.E. = -75% <---- DONE...

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Or you can simply apply the formula :->
---> $\frac{K_{initial}}{K_{final}} = \frac{ u^{2} }{ v^{2} }$

Hence... --> $\frac{K_{initial}}{K_{final}} = \frac{ u^{2} }{ [1/2u]^{2} }$ = 4/1

Therefore, New K.E. = 1/4 times Initial K.E

_________☺☺☺☺ _______________☺☺☺☺________

But it is always advisable to avoid easy method.... Enjoy xD