1. ## Work and Integrals

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

I don't know how to set up an equation for this one, that's pretty much what I need help with for this problem.

2) A tank full of water. Find the work required to pump the water out of the spout.

2. x^3+x^2-2x
As always in physics, the first step is to convert to SI units.
4lb = 1.814 kg
40lb = 18.14 kg
80ft = 24.38 m
2ft/sec = .6096 m/s
.2lb/sec = .09072 kg/s

W = mgh.
we can change this into
$\displaystyle W = \int_{0}^{24.38} mgdh$.

We also know the following:
$\displaystyle \frac {dh}{dt} = .6096 m/s$
$\displaystyle \frac {dm}{dt} = -.09072 kg/s$
$\displaystyle m_0 = 18.14+1.814 = 19.96kg$

Solving for m in terms of t gives us m = 19.96-.09072t
Solving for dh gives dh = .6096 dt
We also need to convert the end points of the interval to times. t=0 is at the bottom of the well and the bucket reaches the top 40sec later so the integral is from 0 to 40

So $\displaystyle W = \int_{0}^{40}9.8(19.96-0.9072t)0.6096 dt$

And your equation is now set up.

2) A tank full of water. Find the work required to pump the water out of the spout.
Again, W= mgh gets turned into $\displaystyle W = \int_{2}^{5}mgdh$.

$\displaystyle \frac {dm}{dh} = 1000\frac {dV}{dh}$ because the density of water is $\displaystyle 1000kg/m^3$
Using similar triangles on either of the ends of the tank tells us that the width of the area on top will always be the same as the depth. Since the depth is 5-h, $\displaystyle \frac {dV}{dh} = 8(5-h)$.
Substituting this in gives $\displaystyle \frac {dm}{dh} = 40 000-8000h$, so $\displaystyle m = 40000h-4000h^2$. This gives an expression for work:
$\displaystyle W = \int_{2}^{5} (40000-4000h)9.8dh$.

and yet another equation is set up.

3. Thank You a lot.