Answer:
[tex]\Delta E_{ham} = 31004.825\,J[/tex]
Explanation:
Let consider that bottom has a height of zero. The change in total energy of the hammerhead is modelled after the Principle of Energy Conservation and Work-Energy Theorem:
[tex]U_{g,A} + \Delta E_{ham} = K_{B}[/tex]
[tex]\Delta E_{ham} = K_{B}-U_{g,A}[/tex]
[tex]\Delta E_{ham} = \frac{1}{2}\cdot m_{ham} \cdot v_{B}^{2}-m_{ham}\cdot g \cdot h_{A}[/tex]
[tex]\Delta E_{ham} = m_{ham}\cdot (\frac{1}{2}\cdot v_{B}^{2}-g\cdot h_{A} )[/tex]
[tex]\Delta E_{ham} = (25\,kg)\cdot [\frac{1}{2}\cdot (50\,\frac{m}{s} )^{2} - (9.807\,\frac{m}{s^{2}} )\cdot (1\,m)][/tex]
[tex]\Delta E_{ham} = 31004.825\,J[/tex]
