The kinetic energy (KE) of an object is given by the formula:
\[ \text{KE} = \frac{1}{2} m v^2 \]
where \( m \) is the mass of the object and \( v \) is its speed.
If the hawk doubles its speed, let's denote its initial speed as \( v \). The initial kinetic energy is:
\[ \text{KE}_{\text{initial}} = \frac{1}{2} m v^2 = 5.1 , \text{J} \]
When the speed is doubled (\( 2v \)), the new kinetic energy becomes:
\[ \text{KE}{\text{final}} = \frac{1}{2} m (2v)^2 = \frac{1}{2} m (4v^2) = 2 \cdot \left( \frac{1}{2} m v^2 \right) = 2 \cdot \text{KE}{\text{initial}} \]
Thus:
\[ \text{KE}_{\text{final}} = 2 \cdot 5.1 , \text{J} = 10.2 , \text{J} \]
Therefore, the best prediction for the hawk's kinetic energy at the end of the chase is 10.2 J.