Let's tackle the problem step by step.

1) Force Applied in Terms of $F(x)$

We are given the following:

• The initial mass of the bucket and sand is 40 kg.
• The bucket leaks sand at a rate of 0.05 kg per meter as it is lifted.
• The height of the building is 25 meters.

Step 1: Determine the mass of the bucket at a height $x$

At any height $x$, the bucket loses sand at a rate of 0.05 kg per meter. Thus, the mass of the bucket as a function of $x$ is given by:

$m(x) = 40 - 0.05x$

where $m(x)$ is the mass (in kg) of the bucket at height $x$ meters.

Step 2: Force due to gravity

The force required to lift the bucket is the weight, which is the mass times the acceleration due to gravity $g = 9.8 \, \text{m/s}^2$.

The force $F(x)$ at height $x$ is:

$F(x) = m(x) \cdot g = (40 - 0.05x) \cdot 9.8$

Thus, the force as a function of $x$ is:

$F(x) = 9.8(40 - 0.05x) = 392 - 0.49x \, \text{N}$

2) Work Done to Lift the Bucket to the Top

The work $W$ is the integral of the force $F(x)$ over the distance $x$ from 0 to 25 meters.

$W = \int_0^{25} F(x) \, dx = \int_0^{25} (392 - 0.49x) \, dx$

Step 1: Compute the integral

The integral of $F(x) = 392 - 0.49x$ is straightforward:

$W = \left[ 392x - \frac{0.49x^2}{2} \right]_0^{25}$

Substitute the limits of integration:

[ W = \left[ 392(25) - \frac{0.49(25^2)}{2} \right] - \left[ 392(0) - \frac{0.49(0^2)}{2} \right] ]

Simplifying:

$W = 392(25) - \frac{0.49(625)}{2}$

$W = 9800 - \frac{0.49 \cdot 625}{2} = 9800 - 153.125$

$W = 9646.875 \, \text{Joules}$

Thus, the work done is approximately:

$W = 9646.875 \, \text{Joules}$

Final Answers:

1. $F(x) = 392 - 0.49x \, \text{N}$
2. $W = 9646.875 \, \text{Joules}$

Would you like any additional details or have any questions?

Here are 5 related questions:

1. What would be the work done if the mass leakage rate was 0.1 kg per meter?
2. How would the force equation change if the height of the building was 50 meters instead?
3. What happens to the force $F(x)$ as the bucket approaches the top of the building?
4. How would the work be affected if the bucket's initial mass was 50 kg?
5. Can you determine the average force applied over the entire lift?

Tip: The work done in lifting an object is directly related to the change in potential energy, and in this case, potential energy decreases as sand is lost.