91Ó°ÊÓ

The given data is listed below as-

• The mass of the Cheetah is $m=70\mathrm{kg}$
• The velocity of the Cheetah is, $V=32.2m/s$

The kinetic energy of a particle equals the amount of work required to accelerate the particle from rest to speed V. Therefore, kinetic energy on the particle is given by-

$\mathbf{K}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{mV}}^{\mathbf{2}}$

The kinetic energy is a scalar and it is always positive or zero.

The kinetic energy of an object is given by-

$K=\frac{1}{2}m{V}^2$

Here, m is the mass of the cheetah, and Vis the velocity of the cheetah.

For,$m=70\mathrm{kg}$ and $V=32.2\mathrm{m}/\mathrm{s}$

Therefore, the Kinetic energy of the cheetah is given by-

role="math" localid="1665210445433" $$\begin{gathered} \mathrm{K}=\frac{1}{2}{\mathrm{mV}}^2 \\ \mathrm{K}=\frac{1}{2}×\left(70\mathrm{kg}\right)×{\left(32.2\mathrm{m}/\mathrm{s}\right)}^2 \\ \mathrm{K}=36.3\mathrm{KJ} \end{gathered}$$

Thus, the Kinetic energy of the cheetah is 36.3 KJ.

The speed of the cheetah is doubled,

Therefore,

$V_2=2V_1$

The new kinetic energy will be

$$\begin{gathered} K.E_2=\frac{1}{2}m{\left(2V_1\right)}^2 \\ K.E_2=2m{V_1}^2 \end{gathered}$$

The ratio of KE₂ and KE₁ is given by-

$$\begin{gathered} \frac{K{E}_2}{K{E}_1}=\frac{2m{V_1}^2}{{\displaystyle \frac{1}{2}}m{V_1}^2} \\ \frac{K{E}_2}{K{E}_1}=4 \\ K{E}_2=4K{E}_1 \end{gathered}$$

Thus, the kinetic energy would change by a factor of 4 when the speed is doubled.