Find the work done in pulling the bucket to the top of the well.

  • Thread starter Thread starter credd741
  • Start date Start date
  • Tags Tags
    Work Work done
credd741
Messages
5
Reaction score
0
A bucket that weighs 5 lb and a rope of negligible weight are used to draw water from a well that is 70 ft deep. The bucket is filled with 50 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.


i have no idea where to start this problem. it is unlike everyothe one i have done so far. help wuld be appreceated.:cry:
 
Physics news on Phys.org
What equations do you know for the work done?
 
the work done is the force times the distance moved.
but I am trying to figure out am i supposed to multiply the weight of the bucket times the the distance per second. or am i supposed to do something else?
 
You'll need to use some properties of differential calculus. For example you can write:

\frac{dm}{dt} = \frac{dm}{dx}\frac{dx}{dt}

Is there any way to write the mass in terms of the distance and work?
 
credd741 said:
the work done is the force times the distance moved.
but I am trying to figure out am i supposed to multiply the weight of the bucket times the the distance per second. or am i supposed to do something else?

Surely, this problem wasn't just dumped on you with no explanation at all! What course is this in, Physics? Algebra? Calculus? Since you posted this under mathematics, I assume it is not a Physics class. If it is in a math class I suspect Calculus and that should tell you something! "Work equals force times distance" is only true if the force is constant over the entire distance. Here the force is the weight of the bucket and water combined and, since water is continually leaking out, that is continually decreasing. What you could do is imagine a very tiny section, \Delta x, and treat the weight as if it were constant over that tiny lift: \Delta Work= F(x)\Delta x where F(x) is the weight at that particular height x. To get an approximate value for the total work you would have to add all of those up. Now what happens to that sum if you take \Delta x going to 0? I suggest you review other examples in this section of your textbook.

And since this is clearly a homework problem, I am moving the thread.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top