To determine the work performed by the force of gravity on a drop that evaporates uniformly, we need to analyze the situation step by step. The drop starts with an initial mass \( M \) and loses mass \( m \) every second until it completely evaporates. The work done by gravity can be calculated using the concept of gravitational force and the distance traveled by the drop during its evaporation.
Understanding the Problem
The drop is subjected to gravitational force, which can be expressed as \( F = mg \), where \( g \) is the acceleration due to gravity. As the drop loses mass, its weight decreases over time. We need to find the total work done by gravity from the moment the drop starts falling until it completely evaporates.
Key Variables
- Initial mass (M): The mass of the drop at the beginning.
- Mass loss rate (m): The mass lost by the drop every second.
- Time (t): The total time until the drop completely evaporates, which can be calculated as \( t = \frac{M}{m} \).
Calculating the Work Done
To find the work done by gravity, we need to consider the distance fallen by the drop at each moment in time. The drop will fall a distance \( h \) under the influence of gravity, which can be calculated using the kinematic equation:
Distance fallen: The distance fallen in time \( t \) is given by \( h = \frac{1}{2} g t^2 \).
However, since the mass of the drop is changing, we need to express the work done in terms of the mass at each moment. The mass of the drop at any time \( t' \) can be expressed as:
Mass at time t': \( M(t') = M - mt' \).
Work Done in Small Time Intervals
The work done by gravity during a small time interval \( dt' \) can be expressed as:
dW = F \cdot dh = (M - mt')g \cdot v dt'
where \( v \) is the velocity of the drop at time \( t' \). The velocity can be expressed as \( v = g t' \) (since it starts from rest). Therefore, we can rewrite the work done as:
dW = (M - mt')g \cdot (g t') dt'
Integrating to Find Total Work
To find the total work done from \( t' = 0 \) to \( t' = \frac{M}{m} \), we need to integrate:
W = ∫(0 to M/m) (M - mt')g \cdot (g t') dt'
This integral can be split into two parts:
- First part: \( gM \int_0^{M/m} t' dt' \)
- Second part: \( -mg \int_0^{M/m} t'^2 dt' \)
Calculating these integrals gives:
- First part: \( gM \cdot \frac{(M/m)^2}{2} = \frac{gM^3}{2m^2} \)
- Second part: \( -mg \cdot \frac{(M/m)^3}{3} = -\frac{gM^3}{3m^2} \)
Combining the Results
Now, we can combine these results to find the total work done:
W = \frac{gM^3}{2m^2} - \frac{gM^3}{3m^2} = gM^3 \left(\frac{1}{2} - \frac{1}{3}\right) \frac{1}{m^2} = gM^3 \cdot \frac{1}{6m^2}
Final Result
Thus, the total work performed by the force of gravity during the time interval from the start to the complete evaporation of the drop is:
W = \frac{gM^3}{6m^2}
This result shows how the work done by gravity depends on the initial mass of the drop and the rate of mass loss, providing insight into the dynamics of falling and evaporating objects under gravity.
