Forces acting on a particle moving in a straight line varies with the velocity of the particle as `F = (alpha)/(upsilon)` where `alpha` is constant. The work done by this force in time interval `Delta t` is :

Explanation

To find the work done by the force acting on a particle moving in a straight line, where the force varies with the velocity of the particle as \( F = \frac{\alpha}{v} \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Relationship Between Force and Power**:
The power \( P \) delivered by a force \( F \) acting on a particle moving with velocity \( v \) is given by:
\[ P = F \cdot v \]

2. **Substitute the Expression for Force**:
Substitute \( F = \frac{\alpha}{v} \) into the power equation:
\[ P = \left(\frac{\alpha}{v}\right) \cdot v = \alpha \]
This shows that the power is constant and equal to \( \alpha \).

3. **Relate Power to Work Done**:
The power can also be expressed as the rate of change of work done \( W \):
\[ P = \frac{dW}{dt} \]
Therefore, we can equate the two expressions for power:
\[ \frac{dW}{dt} = \alpha \]

4. **Integrate to Find Work Done**:
To find the total work done \( W \) over a time interval \( \Delta t \), we integrate both sides:
\[ W = \int_0^{\Delta t} \alpha \, dt \]

5. **Evaluate the Integral**:
The integral of a constant \( \alpha \) over the time interval from \( 0 \) to \( \Delta t \) is:
\[ W = \alpha \int_0^{\Delta t} dt = \alpha [t]_0^{\Delta t} = \alpha (\Delta t - 0) = \alpha \Delta t \]

6. **Final Result**:
Thus, the work done by the force in the time interval \( \Delta t \) is:
\[ W = \alpha \Delta t \]