A hawk is chasing a sparrow. At the beginning of the chase, the hawk’s kinetic energy is 5.1 Joules. At the end of the chase, the hawk doubles its speed. What is the best prediction for its kinetic energy at the end of the chase?(1 point)

Responses

2.55 J
2.55 J

20.4 J
20.4 J

5.1 J
5.1 J

10.2 J

1 answer

The kinetic energy (KE) of an object is given by the formula:

\[ KE = \frac{1}{2} mv^2 \]

where \(m\) is mass and \(v\) is speed. If the hawk doubles its speed, the new kinetic energy can be predicted by considering how kinetic energy changes with speed.

If the initial speed is \(v\), the initial kinetic energy is:

\[ KE_{\text{initial}} = \frac{1}{2} m v^2 = 5.1 , \text{J} \]

When the speed is doubled, the new speed is \(2v\), and the new kinetic energy becomes:

\[ KE_{\text{final}} = \frac{1}{2} m (2v)^2 = \frac{1}{2} m (4v^2) = 4 \left(\frac{1}{2} mv^2\right) = 4 \times KE_{\text{initial}} \]

Substituting the initial kinetic energy:

\[ KE_{\text{final}} = 4 \times 5.1 , \text{J} = 20.4 , \text{J} \]

Thus, the best prediction for the hawk's kinetic energy at the end of the chase is 20.4 J.