A block of mass 2.00 kg moving at a speed of 10.0 m/s accelerates at 3.0 `m/s^2` for 5.00 s. Compute its final kinetic enegy.

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Identify the given values
- Mass of the block (m) = 2.00 kg
- Initial speed (u) = 10.0 m/s
- Acceleration (a) = 3.0 m/s²
- Time (t) = 5.00 s

### Step 2: Calculate the final velocity after 5 seconds
We will use the formula for final velocity under constant acceleration:
\[ v = u + at \]
Substituting the known values:
\[ v = 10.0 \, \text{m/s} + (3.0 \, \text{m/s}^2 \times 5.00 \, \text{s}) \]
\[ v = 10.0 \, \text{m/s} + 15.0 \, \text{m/s} \]
\[ v = 25.0 \, \text{m/s} \]

### Step 3: Calculate the final kinetic energy
The formula for kinetic energy (KE) is:
\[ KE = \frac{1}{2} mv^2 \]
Substituting the values of mass and final velocity:
\[ KE = \frac{1}{2} \times 2.00 \, \text{kg} \times (25.0 \, \text{m/s})^2 \]
Calculating \( (25.0 \, \text{m/s})^2 \):
\[ (25.0 \, \text{m/s})^2 = 625 \, \text{m}^2/\text{s}^2 \]
Now substituting this back into the kinetic energy formula:
\[ KE = \frac{1}{2} \times 2.00 \, \text{kg} \times 625 \, \text{m}^2/\text{s}^2 \]
\[ KE = 1.00 \, \text{kg} \times 625 \, \text{m}^2/\text{s}^2 \]
\[ KE = 625 \, \text{J} \]

### Final Answer
The final kinetic energy of the block after 5 seconds is **625 Joules**.

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