# Integration: Applications to Physics and Engineering

1. Sep 22, 2010

1. The problem statement, all variables and given/known data
A water container is in the shape of a box with a 1-foot square base and height 2-feet. Alas, the container has a hole half way up its side. Fortunately the hole is corked to prevent leaking. The container is filled with water and lifted 100-feet at a rate of 1/2 foot per second by a 100 foot rope weighing 20 lbs. Twenty seconds into the lift, the cork pops out and water leaks from the container at 0.5 lb/sec. How much work is done in the 100 foot lift?

2. Relevant equations
d = rt
w = fd
w =[0-100]\int[F(x)dx]

3. The attempt at a solution
Here is what I have done so far:
volume of box = 2*1*1 = 2ft^3
weight of water: m = dv => 62.5lbs/ft^3 * 2ft^3 == 125lbs (box with water intially)
rope weight: 20lbs
Force intial: 145lbs
water loss: D(weight)/dt = -0. 5 lb/sec
so F(t) = 145 - 0.5t

To get t:
D = rt => rate of ascent * time => 0.5 ft/sec * t = x
t = 2x

so substituting: F(t) = 145 - 0.5t, I got F(x) = 145 - 0.5(2x) == 145 - x

Now, integrating:
w = integral (from 0 to 100) 145 - x dx

w = 145x - (1/2)* x^2 ] 0 to 100

which gives: w = 14500 - 5000 => 9500 ft/lb == answer?

PLEASE SOMEBODY LET ME KNOW IF MY PROCESS IS CORRECT OR NOT. I WOULD HATE TO THINK THAT I'M BEGINNING TO UNDERSTAND THIS PROCESS AND IN REALITY DOING IT ALL WRONG.

2. Sep 22, 2010

Anybody? I would truly appreciate it if someone could critic my work...