A particle of mass m is lying on smooth horizontal table. A constant force F tangential to the surface is applied on it. Find .
(a) average power over a time interval from `t=0` to `t=t,`
(b) instantaneous power as function of time t.

Explanation

To solve the problem, we will break it down into two parts: (a) finding the average power over a time interval from \( t = 0 \) to \( t = t \), and (b) finding the instantaneous power as a function of time \( t \).

### Part (a): Average Power

1. **Identify the Given Information**:
- Mass of the particle: \( m \)
- Constant force applied: \( F \)
- Time interval: from \( t = 0 \) to \( t = t \)

2. **Determine the Acceleration**:
- According to Newton's second law, the acceleration \( a \) of the particle can be calculated as:
\[ a = \frac{F}{m} \]

3. **Calculate the Velocity at Time \( t \)**:
- Since the particle starts from rest (\( u = 0 \)), we can use the equation of motion:
\[ v = u + at \]
- Substituting the values, we get:
\[ v = 0 + \left(\frac{F}{m}\right) t = \frac{F t}{m} \]

4. **Calculate the Work Done**:
- The work done \( W \) on the particle is equal to the change in kinetic energy. The kinetic energy \( KE \) at time \( t \) is given by:
\[ KE = \frac{1}{2} mv^2 \]
- Substituting the expression for \( v \):
\[ KE = \frac{1}{2} m \left(\frac{F t}{m}\right)^2 = \frac{1}{2} m \cdot \frac{F^2 t^2}{m^2} = \frac{F^2 t^2}{2m} \]

5. **Calculate the Average Power**:
- Average power \( P_{\text{avg}} \) is defined as the work done divided by the time taken:
\[ P_{\text{avg}} = \frac{W}{t} = \frac{\frac{F^2 t^2}{2m}}{t} = \frac{F^2 t}{2m} \]

### Part (b): Instantaneous Power

1. **Define Instantaneous Power**:
- The instantaneous power \( P_I \) is given by the formula:
\[ P_I = F \cdot v \cdot \cos(\theta) \]
- Since the force is applied tangentially to the surface, \( \theta = 0 \) degrees, and thus \( \cos(0) = 1 \):
\[ P_I = F \cdot v \]

2. **Substitute the Expression for Velocity**:
- We already found that:
\[ v = \frac{F t}{m} \]
- Substituting this into the power equation gives:
\[ P_I = F \cdot \left(\frac{F t}{m}\right) = \frac{F^2 t}{m} \]

### Final Answers:
- (a) Average Power:
\[ P_{\text{avg}} = \frac{F^2 t}{2m} \]
- (b) Instantaneous Power:
\[ P_I = \frac{F^2 t}{m} \]