# Thread: Emptying a pool

1. ## Emptying a pool

Hello Mathhelpforum!
I have two questions. One with an answer I wanted to double check, and the second needing help on.

The first one is if it takes 100J of work to stretch a spring 0.5m from it's equilibrium position, how much work is needed to stretch an additional .75m?

So I set my limits at x = .5 and x = 1.25.
100 = F(.5)
F = 200.
kx = F
k(.5) = 200
k = 100.
So the spring constant is 100. Then the integral of 100x dx from .5 to 1.25 = 50x² -> 5(1.25)² - (50(.5)²) = 65.625J

And this problem I'm stuck on.
A swimming pool has a shape of a box with a base of 25m by 15m and a depth of 2.5m. How much work is required to pump the water out of the pool?
Would it be Integral from 0 to 2.5m (pg)dy * integral from 0 to 2.5m of (2.5-y)dy?

2. ## Re: Emptying a pool

Some error here I think.
Extending from 0 to 0.5 work done= integral of kx from 0 to 0.5=k(0.5)^2/2 So 0.25k/2=100 k=800

3. ## Re: Emptying a pool

Hooke's law says f = -kx so k = f/x?

4. ## Re: Emptying a pool

Originally Posted by Verleiren
Hooke's law says f = -kx so k = f/x?
the force is variable ... you're trying to treat it as a constant.

$\displaystyle W = \frac{1}{2}kx^2$

$\displaystyle 100 = \frac{1}{2}k(0.5)^2$ , solve for k, then calculate $\displaystyle \int_{0.5}^{1.25} kx \, dx$

for the pool ...

$\displaystyle W = \rho \cdot g \int_0^{2.5} (25 \cdot 15)(2.5 - y) \, dy$ , where $\displaystyle \rho$ is the density of water and $\displaystyle g$ is the acceleration due to gravity