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)

Question

Responses

10.2 J
10.2 J

2.55 J
2.55 J

20.4 J
20.4 J

5.1 J

Answer 1

Kinetic energy (KE) is given by the formula:

$$KE=\frac{1}{2}m{v}^2$$

where $m$ is the mass and $v$ is the velocity.

If the hawk doubles its speed, then its new kinetic energy will be:

$$K{E}_{\text{new}}=\frac{1}{2}m(2v{)}^2=\frac{1}{2}m(4v^2)=4\left(\frac{1}{2}m{v}^2\right)=4\times K{E}_{\text{initial}}$$

Given that the initial kinetic energy is 5.1 Joules:

$$K{E}_{\text{new}}=4\times 5.1,\text{J}=20.4,\text{J}$$

So, the best prediction for the hawk's kinetic energy at the end of the chase is:

20.4 J