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

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


Sagot :

Let's solve this problem step-by-step using the kinetic energy equation \( KE = \frac{1}{2}mv^2 \).

Given:
- Mass of the puppy, \( m = 3 \) kilograms
- Initial velocity, \( v_{\text{initial}} = 2 \) meters per second
- Final velocity, \( v_{\text{final}} = 1 \) meter per second

### Step 1: Calculate Initial Kinetic Energy
The initial kinetic energy can be calculated using the initial velocity:
[tex]\[ KE_{\text{initial}} = \frac{1}{2} \times m \times (v_{\text{initial}})^2 \][/tex]

Substitute the given values:
[tex]\[ KE_{\text{initial}} = \frac{1}{2} \times 3 \times (2)^2 = \frac{1}{2} \times 3 \times 4 = \frac{1}{2} \times 12 = 6 \text{ Joules} \][/tex]

### Step 2: Calculate Final Kinetic Energy
The final kinetic energy can be calculated using the final velocity:
[tex]\[ KE_{\text{final}} = \frac{1}{2} \times m \times (v_{\text{final}})^2 \][/tex]

Substitute the given values:
[tex]\[ KE_{\text{final}} = \frac{1}{2} \times 3 \times (1)^2 = \frac{1}{2} \times 3 \times 1 = \frac{1}{2} \times 3 = 1.5 \text{ Joules} \][/tex]

### Step 3: Calculate the Change in Kinetic Energy
The change in kinetic energy is given by the difference between the final and initial kinetic energies:
[tex]\[ \Delta KE = KE_{\text{final}} - KE_{\text{initial}} \][/tex]

Substitute the calculated values:
[tex]\[ \Delta KE = 1.5 - 6 = -4.5 \text{ Joules} \][/tex]

### Conclusion
Since the change in kinetic energy is \(-4.5\) Joules, this means that the puppy's kinetic energy decreases. The final kinetic energy of the puppy is \(1.5\) Joules.

Therefore, the correct answer is:
A. Her kinetic energy decreases to [tex]$1.5 J$[/tex].