To solve this problem, we calculate the work done lifting the bucket with the mass decreasing as the sand leaks out. Here's the detailed solution:

Given:

1. Initial mass of bucket and sand = $15 \, \text{kg}$
2. Height = $25 \, \text{m}$
3. Rate of sand leakage = $0.2 \, \text{kg/m}$
4. Gravitational acceleration = $g = 9.8 \, \text{m/s}^2$

Step-by-step process:

1. Mass as a function of height

The bucket loses $0.2 \, \text{kg}$ for every meter lifted. Hence, the mass $m(h)$ at a height $h$ is:

$m(h) = 15 - 0.2h$

2. Work done lifting the bucket

The work done $W$ in lifting an object is given by:

$W = \int_0^{25} m(h) \cdot g \, dh$

Substitute $m(h) = 15 - 0.2h$:

$W = \int_0^{25} (15 - 0.2h) \cdot 9.8 \, dh$

$W = 9.8 \int_0^{25} (15 - 0.2h) \, dh$

3. Solve the integral

Expand and integrate:

$W = 9.8 \left[ \int_0^{25} 15 \, dh - \int_0^{25} 0.2h \, dh \right]$

$W = 9.8 \left[ 15 \cdot h \big|_0^{25} - 0.2 \cdot \frac{h^2}{2} \big|_0^{25} \right]$

Evaluate each term:

• $\int_0^{25} 15 \, dh = 15 \cdot 25 - 15 \cdot 0 = 375$
• $\int_0^{25} 0.2h \, dh = 0.2 \cdot \frac{25^2}{2} - 0.2 \cdot \frac{0^2}{2} = 0.2 \cdot 312.5 = 62.5$

Substitute back: $W = 9.8 \left[ 375 - 62.5 \right]$

$W = 9.8 \cdot 312.5$

$W = 3062.5 \, \text{J}$

Final Answer:

The work done lifting the bucket is 3062.5 Joules.

Would you like more clarification on any step? Here are related questions you might find helpful:

1. How does sand leakage affect the total work done?
2. What is the physical significance of integrating mass as a function of height?
3. How would the answer change if the sand leaked at a different rate?
4. How can this problem be visualized with a graph of mass vs height?
5. Can we derive an alternative method without calculus?

Tip: Always consider variable mass problems using integrals to account for gradual changes systematically.