91Ó°ÊÓ

The expression for the Kinetic Energy is given by,

$\mathit{K}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}^{\mathbf{2}}$

Here, mis the mass of an object, and v is the velocity of the object.

The brick is moving in a downward direction. The velocity of the brick is .

$\left({\stackrel{→}{v}}_b\right)=-5.0\mathrm{m}/\mathrm{s}$

The velocity of the melon is $\left({\stackrel{→}{v}}_m\right)=3.0\hat{i}+4.0\hat{j}$. Here,$\hat{i}$ , and$\hat{j}$ are unit vectors along the horizontal and vertical direction respectively.

The magnitude of the velocity is,

$$\begin{gathered} \left|{\stackrel{→}{v}}_m\right|=\sqrt{{\left(3\right)}^2+{\left(4\right)}^2} \\ =\sqrt{25} \\ =5\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the velocities are not the same.

Calculate the kinetic energy of the brick.

$$\begin{gathered} K_b=\frac{1}{2}m{v}_b^2 \\ =\frac{1}{2}\left(1.5\mathrm{kg}\right){\left(-5\mathrm{m}/\mathrm{s}\right)}^2 \\ =18.75\mathrm{J} \end{gathered}$$

Calculate the kinetic energy of the physics book.

$$\begin{gathered} K_p=\frac{1}{2}m{v}_p^2 \\ =\frac{1}{2}\left(1.5\mathrm{kg}\right){\left(5\mathrm{m}/\mathrm{s}\right)}^2 \\ =18.75\mathrm{J} \end{gathered}$$

Calculate the kinetic energy of the melon.

$$\begin{gathered} K_m=\frac{1}{2}m{v}_m^2 \\ =\frac{1}{2}\left(1.5\mathrm{kg}\right){\left(5\mathrm{m}/\mathrm{s}\right)}^2 \\ =18.75\mathrm{J} \end{gathered}$$

Therefore, the kinetic energy of all bodies are the same.