A block of mass 1kg start moving with constant acceleration `a=4m//s_(2)` Find.
(a) average power of the net force in time inteval from `t=0` to `t=2s`,
(b) instantaneous power of the net force at `t=4 s`.

Explanation

To solve the given problem, we will break it down into two parts: (a) finding the average power of the net force from \( t = 0 \) to \( t = 2 \) seconds, and (b) finding the instantaneous power of the net force at \( t = 4 \) seconds.

### Part (a): Average Power Calculation

1. **Identify the given values:**
- Mass of the block, \( m = 1 \, \text{kg} \)
- Acceleration, \( a = 4 \, \text{m/s}^2 \)
- Time interval, \( t = 2 \, \text{s} \)

2. **Calculate the final velocity at \( t = 2 \) seconds:**
- Using the formula for final velocity under constant acceleration:
\[ v = u + at \]
- Here, the initial velocity \( u = 0 \) (the block starts from rest), so:
\[ v = 0 + (4 \, \text{m/s}^2)(2 \, \text{s}) = 8 \, \text{m/s} \]

3. **Calculate the change in kinetic energy:**
- The change in kinetic energy (\( \Delta KE \)) is given by:
\[ \Delta KE = \frac{1}{2} m v^2 - \frac{1}{2} m u^2 \]
- Substituting the values:
\[ \Delta KE = \frac{1}{2} (1 \, \text{kg}) (8 \, \text{m/s})^2 - \frac{1}{2} (1 \, \text{kg}) (0 \, \text{m/s})^2 \]
- This simplifies to:
\[ \Delta KE = \frac{1}{2} (1) (64) - 0 = 32 \, \text{J} \]

4. **Calculate the average power:**
- Average power (\( P_{\text{avg}} \)) is given by:
\[ P_{\text{avg}} = \frac{\Delta KE}{\Delta t} \]
- Substituting the values:
\[ P_{\text{avg}} = \frac{32 \, \text{J}}{2 \, \text{s}} = 16 \, \text{W} \]

### Part (b): Instantaneous Power Calculation

1. **Calculate the instantaneous power at \( t = 4 \) seconds:**
- Instantaneous power (\( P \)) is given by:
\[ P = F \cdot v \]
- Where \( F \) (net force) can be calculated using Newton's second law:
\[ F = m \cdot a = 1 \, \text{kg} \cdot 4 \, \text{m/s}^2 = 4 \, \text{N} \]

2. **Calculate the velocity at \( t = 4 \) seconds:**
- Using the same formula for final velocity:
\[ v = u + at = 0 + (4 \, \text{m/s}^2)(4 \, \text{s}) = 16 \, \text{m/s} \]

3. **Substitute the values into the instantaneous power formula:**
- Thus, the instantaneous power is:
\[ P = F \cdot v = 4 \, \text{N} \cdot 16 \, \text{m/s} = 64 \, \text{W} \]

### Final Answers:
- (a) Average Power: \( 16 \, \text{W} \)
- (b) Instantaneous Power: \( 64 \, \text{W} \)