Question

Calculate the change in the kinetic energy of a moving body if its velocity is reduced to 1/3rd of the initial velocity.

Answer 1

Hint:
Kinetic energy is a form of energy that an object or a particle has by reason of its motion. If work, which transfers energy, is done on an object by applying a net force, the object speeds up and thereby gains kinetic energy.

Complete step-by-step answer:
Let, the initial velocity is ${v_i}$ and final velocity ${v_f}$
According to the question final velocity is reduced to 1/3 rd of the initial velocity
So, ${v_f} = \dfrac{1}{3}{v_i}$ ...............equation (1)
According to the kinetic energy formula
We know,
$K = \dfrac{1}{2}m{v^2}$ ...........equation (2)
Now the initial kinetic energy is
 ${K_i} = \dfrac{1}{2}m{v_i}^2$.........equation 3
And final kinetic velocity is,
${K_f} = \dfrac{1}{2}m{v_f}^2$
Now put the value of ${v_f}$ from equation 1
We get,
$ \Rightarrow $${K_f} = \dfrac{1}{2}m\dfrac{{{v_i}^2}}{9}$
Now, from the equation 3
We get,
${K_f} = \dfrac{{{K_i}}}{9}$..............because${K_i} = \dfrac{1}{2}m{v_i}^2$
Now change of kinetic energy is
$\begin{gathered}
  \therefore \Delta K = {K_i} - {K_f} \\
   \Rightarrow \Delta K = {K_i} - \dfrac{{{K_i}}}{9} \\
   \Rightarrow \Delta K = \dfrac{{8{K_i}}}{9} \\
\end{gathered} $
So the change of kinetic energy is equal to 8/9 th time of initial kinetic energy.

Note:
Kinetic energy is directly proportional to the mass of the object and to the square of its velocity: $K = \dfrac{1}{2}m{v^2}$. If the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared.