Given data
*The given mass of the rock is m = 2 kg
*The given potential energy is U_p = 407 J
(a)
The diagram of the energy bar graph is drawn below
(b)
If an object is at rest and has potential energy, once it starts to fall from its rest state then this potential energy is completely transferred to kinetic energy. This means that the magnitude of the kinetic energy is equal to the potential energy of the object.
The change in kinetic energy of the rock while falling to the ground is given as
[tex]\begin{gathered} U_k=U_p \\ =407\text{ J} \end{gathered}[/tex]
(c)
The formula for the speed of the block is given as
[tex]\begin{gathered} U_k=\frac{1}{2}mv^2 \\ v=\sqrt[]{\frac{2U_k}{m}} \end{gathered}[/tex]
Substitute the known values in the above expression as
[tex]\begin{gathered} v=\sqrt[]{\frac{2\times407}{2}} \\ =20.17\text{ m/s} \end{gathered}[/tex]
Hence, the speed of the object is v = 20.17 m/s
