Solving the Hardest Work Problem Homework

[derek181]

Homework Statement

A bucket of water with mass 100kg is on the ground attached to one end of a cable with mass per unit length of 5kg/m. The other end of the cable is attached to a windlass 100m above the bucket. if the bucket is raised at a constant speed, water runs out through a hole in the bottom at a constant rate to the extent that the bucket would have mass 80kg when it reaches the top. To further complicate matters, a pigeon of mass 2kg lands on the bucket when it is 50m above the ground. He immediately begins taking a bath, splashing water over the side of the bucket at the rate of 1kg/m. Find the work done by the windlass in raising the bucket 100m.

Homework Equations

W=FD
∫Fydy

The Attempt at a Solution

No idea, Can someone help me solve this?

 

[derek181]

Ah here is an attempt.

∫¹⁰⁰₀(0.2kg/m)(9.81m/s²)(100-y)dy+∫¹⁰⁰₀(100kg)(9.81m/s²)(100-y)dy+∫¹⁰⁰₅₀(2kg)(9.81m/s²)(100-y)dy+∫¹⁰⁰₅₀(1kg/m)(9.81m/s²)(100-y)dy

 

[Ray Vickson]

Homework Statement

A bucket of water with mass 100kg is on the ground attached to one end of a cable with mass per unit length of 5kg/m. The other end of the cable is attached to a windlass 100m above the bucket. if the bucket is raised at a constant speed, water runs out through a hole in the bottom at a constant rate to the extent that the bucket would have mass 80kg when it reaches the top. To further complicate matters, a pigeon of mass 2kg lands on the bucket when it is 50m above the ground. He immediately begins taking a bath, splashing water over the side of the bucket at the rate of 1kg/m. Find the work done by the windlass in raising the bucket 100m.

Homework Equations

W=FD
∫Fydy

The Attempt at a Solution

No idea, Can someone help me solve this?

You are dealing with a variable-mass problem. See, eg., http://en.wikipedia.org/wiki/Variable-mass_system or
http://physics.stackexchange.com/questions/53980/second-law-of-Newton-for-variable-mass-systems
or http://vixra.org/pdf/1309.0210v1.pdf . These can be quite tricky.

 

[derek181]

I looked at those links and I think you may be overcomplicating the problem. For the parts that are decreasing I found out that I have to take the initial weight minus the rate of decrease multiplied by the length (y). So (100kg-0.2kg/m(ym))dy. I get the correct order of magnitude but the answer is slightly off.

 

[derek181]

This problem is a type where you can solve by evaluating the definite integral.

 

[haruspex]

Ah here is an attempt.

∫¹⁰⁰₀(0.2kg/m)(9.81m/s²)(100-y)dy+∫¹⁰⁰₀(100kg)(9.81m/s²)(100-y)dy+∫¹⁰⁰₅₀(2kg)(9.81m/s²)(100-y)dy+∫¹⁰⁰₅₀(1kg/m)(9.81m/s²)(100-y)dy

I don't seem to be able to match those integrals up with the various components, perhaps because some are erroneous.
Please separate out and identify each of the work components and state the integral for each.