Question

Calculate the ke of a moving body if its velocity is reduced to 1/4 of the initial velocity

Answer 1

Given:

Velocity of a moving body is reduced to 1/4 of the initial velocity

To Find:

Kinetic energy of given body after reducing velocity

Solution:

Let the mass of given body be 'm' , initial velocity be 'u' and initial Kinetic energy of body be K.

We know that,

$\pink{\boxed{\bf{Kinetic\:Energy=\frac{1}{2}\times m\times v^{2}}}}$

where,

m is the mass of body

v is the velocity of body

So,

$\longrightarrow\mathrm{K=\dfrac{1}{2}\times m\times u^{2}}$

$\longrightarrow\mathrm{K=\dfrac{mu^{2}}{2}}--------(1)$

Now, let final velocity be 'v' and final Kinetic energy be K'

Now, according to question

$\longrightarrow\mathrm{v=\dfrac{u}{4}}$

So,

$\longrightarrow\mathrm{K'=\dfrac{1}{2}\times m\times v^{2}}$

$\longrightarrow\mathrm{K'=\dfrac{1}{2}\times m\times \bigg(\dfrac{u}{4}\bigg)^{2}}$

$\longrightarrow\mathrm{K'=\dfrac{1}{2}\times m\times \dfrac{u^{2}}{16}}$

$\longrightarrow\mathrm{K'=\dfrac{mu^{2}}{32}}$

$\longrightarrow\mathrm{K'=\dfrac{1}{16}\bigg(\dfrac{mu^{2}}{2}\bigg)}$

From (1), we get

$\longrightarrow\mathrm{K'=\dfrac{1}{16}\bigg(K\bigg)}$

$\longrightarrow\mathrm{K'=\dfrac{K}{16}}$

Hence, final Kinetic energy is 1/16 times of initial Kinetic energy.