Applications of Integration - Work.

In summary: Use the value of g in appropriate units}= -28224 jouleIn summary, the conversation discusses the concept of
  • #1
JohnRV5.1
8
0
I am currently taking CALC II in the summer. So far we have gone over Applications of Integration(area under the curve, solids of revolution) to Techniques of Integration (integration by parts, Trig substitution, Partial Fractions, IMproper Integrals, etc.) From all the problems I have encountered, only this one gives me trouble. It is the chapter on applications of Integration that pertains to work. Here is the problem and the solution I obtained, altough I am unsure I tackled the problem correctly:

Problem:
A bucket that weights 4 lb and a rope of negligible weight are used to draw water from a well that is 80 ft deep. The bucket is filled with 40 lb of water and is pulled up at a rate of 2 ft/s, but water leaks out of a hole in the bucket at a rate of 0.2 lb/s. Find the work done in pulling the bucket to the top of the well.

My Solution:
First I found the work required to lift the bucket byself to the top of the well.
I got Force = (4 lb)(80 ft) = 320 ft*lb

Then I obtained the work done in pulling the leaking water to the top of the well using integration.
I found the distance = x
The Force = (40 lb)/(80 ft) - (.2 lb/s)/(2 ft/s) = .5 lb/ft - .1lb/ft = .4 lb/ft or 2/5 lb/ft
therefor the force is = 2/5 dx or .4dx.
I set up the integral using the above info
= Integral from 0 to 80 of .4xdx
evaluating the integral I obtained 1280 ft*lb

So I summed up both the work required to lift the bucket to the top and the work required to lift the water to the top of the well: 1280 ft*lb + 320 ft*lb = Work = 1600 ft*lb! Am I correct

p.s. My calc instructor informed us that no work questions will be on the exam but he did assign homework for it. The problem was due to time constraints, the instructor was not able to lecture on the section on work. Plus I have never taken a single Physics class so I did not feel too confident with my solution.
Thank you for lending your time and efforts to help me.
 
Physics news on Phys.org
  • #2
I attempted the problem a second time with a different approach and got 3200 ft*lb of work instead. This time I used the integral from 0 to 80 of
(40 - .1x)dx

Can anyone tell me if I obtained the right solution. BTW, sorry if your having difficulties reading my notation above. Help!
 
  • #3
One way to do this question.
At time 't' what is the mass of water left in the bucket? It is 40 - 0.2t lb

The mass of the bucket is 4lb. So the total mass is 40 -0.2t + 4 = 44 - 0.2t lb.
now, by Newton's second law,

d(mv)/dt=F(net).

Therefore, mdv/dt+vdm/dt=F(net)

Now, dv/dt=0

so, vdm/dt=(T-mg)
where T is the force you apply to pull the bucket as a function of time.
So T= vdm/dt+mg.
T=(2)(-0.2) + (44-0.2t)(g) {Use the value of g in appropriate units}
and time taken to reach the top is h/v=(80)/(2)=40

So work done is
[tex] \int T.dx [/tex]
Which is [tex] \int T.(vdt) [/tex]
= [tex] \int T.(2)(dt) [/tex] from t=0 to t=40
 
Last edited:

What is the concept of work in the context of integration?

The concept of work in the context of integration refers to the calculation of the amount of energy required to move an object a certain distance. This is done by finding the area under a curve on a graph representing a force function.

What are some real-world applications of integration in work?

Some real-world applications of integration in work include calculating the amount of energy required to move an object against a resistive force, such as friction or air resistance. This can be useful in fields such as engineering and physics.

How is the work formula derived from integration?

The work formula, W = F * d, is derived from the fundamental theorem of calculus, which states that the definite integral of a function is equal to the difference between the values of the function at the upper and lower limits of integration. In this case, the force function is integrated with respect to distance, resulting in the work formula.

What is the relationship between work and kinetic energy?

The relationship between work and kinetic energy is given by the work-energy theorem, which states that the net work done on an object is equal to the change in its kinetic energy. This means that the work done on an object is responsible for changing its speed or direction of motion.

How is integration used to calculate the work done by a varying force?

In cases where the force acting on an object varies, integration is used to calculate the work done by finding the area under a curve on a force vs. distance graph. This allows for the determination of the total work done, even when the force is not constant.

Similar threads

  • Introductory Physics Homework Help
Replies
15
Views
2K
  • Introductory Physics Homework Help
Replies
15
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
3K
  • Introductory Physics Homework Help
Replies
11
Views
4K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
3K
  • Introductory Physics Homework Help
Replies
7
Views
10K
  • Introductory Physics Homework Help
2
Replies
37
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
1K
Back
Top