Calc II word problem(work needed to lift a leaky bucket)

  • Thread starter nlsherrill
  • Start date
  • Tags
    Lift
In summary, the problem involves a leaky 10-kg bucket being lifted to a height of 12 meters at a constant speed with a rope that weighs 0.8kg/m. The bucket initially contains 36kg of water, but the water leaks at a constant rate and finishes draining just as the bucket reaches the 12-meter level. The total weight of the water/bucket/rope initially is 55.6 kg, and at the end, only the 10kg bucket is left. The work done can be calculated using the equation W = \int\limits_{a}^{b} F\,dr, where F = mg and m = 10 + 36 + (.8)(12) -
  • #1
nlsherrill
323
1

Homework Statement


From Stewart Calculus Concepts and Contexts 4th edition pg.473 section 6.6 #15...

:A leaky 10-kg bucket is lifted from the ground to a height of 12 meters
at a constant speed with a rope that weighs 0.8kg/m. Initially the bucket contains 36kg of water but the water leaks at a constant rate and finishes draining just as the bucket reaches the 12 meter level. How much work is done

Homework Equations






The Attempt at a Solution



I am able to do these kind of problems, but the only thing different about this one is the weight is constantly changing. The total weight of the water/bucket/rope initially is 55.6 kg, and at the end its just the 10 kg bucket left.

I don't know how to approach the change in weight of the water leaking out of the bucket. So far I think you integrate from 0-12 meters, and the distance an arbitrary part of the rope has to travel is 12-x. Any hints?
 
Physics news on Phys.org
  • #2
[tex] W = \int\limits_{a}^{b} F\,dr[/tex]
[tex]F = mg[/tex]
[tex]m = 10 + 36 + (.8)(12) - .8r - \frac{36r}{12}[/tex]
[tex]a = 0[/tex]
[tex]b = 12[/tex]
[tex]g = 9.8[/tex]
 
  • #3
Ah that's actually quite simple. I thought a new variable would have to be introduced to account for a rate of change. Thanks a lot.
 
  • #4
xcvxcvvc said:
[tex] W = \int\limits_{a}^{b} F\,dr[/tex]
[tex]F = mg[/tex]
[tex]m = 10 + 36 + (.8)(12) - .8r - \frac{36r}{12}[/tex]
[tex]a = 0[/tex]
[tex]b = 12[/tex]
[tex]g = 9.8[/tex]

That makes an assumption about how the rope is routed...

nlsherrill -- can you tell us what assumption he is making about how the rope is routed, and what would be different if it were routed differently?
 
  • #5
berkeman said:
That makes an assumption about how the rope is routed...

nlsherrill -- can you tell us what assumption he is making about how the rope is routed, and what would be different if it were routed differently?

Hmm..well there are a bunch of factors that must be taken into consideration I suppose...but as far as the routing of the rope, I imagine it is fixed and stays in a straight line as it is pulled up 12 meters. Otherwise the amount of force required to keep the bucket moving at a constant rate would fluctuate because the bucket would move some in the x or z directions instead of just strictly the y direction. I believe this is the assumption they made, that the bucket travels strictly in a straight line, and the ropes shape does not change.

I don't know if that's what your looking for, but from the looks of the assumptions they made this problem about as easy as they could(besides the changing of the weight with respect to the distance traveled).
 
  • #6
nlsherrill said:
Hmm..well there are a bunch of factors that must be taken into consideration I suppose...but as far as the routing of the rope, I imagine it is fixed and stays in a straight line as it is pulled up 12 meters. Otherwise the amount of force required to keep the bucket moving at a constant rate would fluctuate because the bucket would move some in the x or z directions instead of just strictly the y direction. I believe this is the assumption they made, that the bucket travels strictly in a straight line, and the ropes shape does not change.

I don't know if that's what your looking for, but from the looks of the assumptions they made this problem about as easy as they could(besides the changing of the weight with respect to the distance traveled).

xcv's solution assumes the rope is pulled up to a ledge or something, and dropped there. How would the equation change if the rope were looped over a single pulley, and pulled from the ground in order to hoist up the bucket...? (that is how it normally would be hoisted anyway, right?)
 
  • #7
berkeman said:
xcv's solution assumes the rope is pulled up to a ledge or something, and dropped there. How would the equation change if the rope were looped over a single pulley, and pulled from the ground in order to hoist up the bucket...? (that is how it normally would be hoisted anyway, right?)

Ha I'm not really sure! I think it would be a different equation altogether because more rope would be needed in that senario. Otherwise the person on the ground wouldn't have a way to pull the bucket up to 12 meters.

Other than that...I don't know. Maybe some signs would change because the direction of action on the person pulling the rope up is different.
 

Related to Calc II word problem(work needed to lift a leaky bucket)

What is the meaning of "work needed to lift a leaky bucket" in Calc II?

In Calc II, the "work needed to lift a leaky bucket" refers to a word problem that involves using integral calculus to calculate the amount of work required to lift a bucket with a leak in it from one point to another.

What is the process for solving a "Calc II word problem(work needed to lift a leaky bucket)"?

To solve a "Calc II word problem(work needed to lift a leaky bucket)", you will first need to set up an integral representing the total work required. This involves finding the weight of the bucket and the distance it needs to be lifted, as well as taking into account the work done by the leak. You will then need to evaluate the integral using appropriate techniques such as substitution or integration by parts.

What are the common mistakes to avoid when solving a "Calc II word problem(work needed to lift a leaky bucket)"?

Some common mistakes to avoid when solving a "Calc II word problem(work needed to lift a leaky bucket)" include forgetting to include the work done by the leak in the integral, not correctly setting up the integral, and making errors during the integration process. It is important to carefully check your work and double-check all calculations to avoid these mistakes.

What are some real-world applications of "Calc II word problem(work needed to lift a leaky bucket)"?

The concept of "work needed to lift a leaky bucket" has many real-world applications, such as calculating the amount of work needed to raise a bucket of water from a well, or the work required to lift a heavy object using a pulley system. It can also be applied in engineering and physics to calculate the amount of energy or force needed to move objects from one point to another.

How can I improve my skills in solving "Calc II word problem(work needed to lift a leaky bucket)"?

To improve your skills in solving "Calc II word problem(work needed to lift a leaky bucket)", it is important to practice regularly and make sure you have a solid understanding of the fundamental principles of integral calculus. You can also seek help from a tutor or attend study groups to get additional support and clarification on any concepts you may be struggling with.

Similar threads

  • Calculus and Beyond Homework Help
Replies
11
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
3K
  • Calculus and Beyond Homework Help
Replies
5
Views
5K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
1K
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
13
Views
9K
  • Introductory Physics Homework Help
Replies
9
Views
5K
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
Back
Top