If the speed of a particle triples, by what factor does its kinetic energy increase?

The object’s kinetic energy can be described as the energy of the particles in motion, observable as the movement of particles from one location to another.

It is directly proportional to the second power of velocity of an object of constant mass.

The equation for the kinetic energy is as follows:

\(K = \frac{1}{2}m{v^2}\)

Here, \(m\) is the mass of the object, and \(v\) is the velocity of the object.

Let \({v_1}\) be the initial velocity of the particle, and \({v_2}\) be the final velocity of the particle.

Since it is given that the speed of the particle is tripled, you can write \({v_2} = 3{v_1}\).

The expression for the initial kinetic energy of the particle can be written as follows:

\({K_1} = \frac{1}{2}mv_1^2\) … (i)

The expression for the final kinetic energy of the particle can be written as follows:

\({K_2} = \frac{1}{2}mv_2^2\) … (ii)

Dividing equation (i) by equation (ii):

\(\begin{aligned}\frac{{{K_1}}}{{{K_2}}} &= \frac{{\frac{1}{2}mv_1^2}}{{\frac{1}{2}mv_2^2}}\\\frac{{{K_1}}}{{{K_2}}} &= \frac{{v_1^2}}{{v_2^2}}\\\frac{{{K_1}}}{{{K_2}}} &= \frac{{v_1^2}}{{{{\left( {3{v_1}} \right)}^2}}}\\\frac{{{K_1}}}{{{K_2}}} &= \frac{1}{9}\\{K_2} &= 9{K_1}\end{aligned}\)

Hence, the kinetic energy of the particle becomes 9 times the initial kinetic energy.