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A puppy weighing 3 kilograms races through the dog park. If she slows from a speed of 2 meters/second to 1 meter/second, what will happen to her kinetic energy? Use the kinetic energy equation [tex][tex]$KE = \frac{1}{2}mv^2$[/tex][/tex].

A. Her kinetic energy decreases to [tex][tex]$1.5 \, J$[/tex][/tex].
B. Her kinetic energy increases to [tex][tex]$1.5 \, J$[/tex][/tex].
C. Her kinetic energy remains unchanged.
D. Her kinetic energy decreases to [tex][tex]$1 \, J$[/tex][/tex].


Sagot :

Let's solve the problem step-by-step:

1. Calculate the initial kinetic energy (KE_initial):
[tex]\[ KE_{\text{initial}} = \frac{1}{2} m v_\text{initial}^2 \][/tex]

- Mass ([tex]\(m\)[/tex]) of the puppy: 3 kilograms
- Initial speed ([tex]\(v_{\text{initial}}\)[/tex]): 2 meters/second

Substituting the values:
[tex]\[ KE_{\text{initial}} = \frac{1}{2} \times 3 \times (2)^2 \][/tex]
[tex]\[ KE_{\text{initial}} = \frac{1}{2} \times 3 \times 4 \][/tex]
[tex]\[ KE_{\text{initial}} = 6 \, J \][/tex]

2. Calculate the final kinetic energy (KE_final):
[tex]\[ KE_{\text{final}} = \frac{1}{2} m v_\text{final}^2 \][/tex]

- Final speed ([tex]\(v_{\text{final}}\)[/tex]): 1 meter/second

Substituting the values:
[tex]\[ KE_{\text{final}} = \frac{1}{2} \times 3 \times (1)^2 \][/tex]
[tex]\[ KE_{\text{final}} = \frac{1}{2} \times 3 \times 1 \][/tex]
[tex]\[ KE_{\text{final}} = 1.5 \, J \][/tex]

3. Calculate the decrease in kinetic energy:
[tex]\[ \text{Decrease in KE} = KE_{\text{initial}} - KE_{\text{final}} \][/tex]

Substituting the calculated values:
[tex]\[ \text{Decrease in KE} = 6 \, J - 1.5 \, J \][/tex]
[tex]\[ \text{Decrease in KE} = 4.5 \, J \][/tex]

From this, we see that her kinetic energy decreases, and the final kinetic energy is [tex]\(1.5 \, J\)[/tex]. Therefore, the correct answer is:

A. Her kinetic energy decreases to [tex]\(1.5 \, J\)[/tex].
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