# Amount of work done (integration)

• February 13th 2008, 04:30 AM
vperera
Amount of work done (integration)
A cable that weighs 2ft/lb is used to lift 800lb of coal up a mine shaft 500ft deep. Find the work done. I can't figure out where to start. I understand that W=Fd.

I also have a problem trying to solve something that seems relatively simple: A particle is moved along the x-axis by a force that measures 10/(1+x)^2 lbs at a point x ft. from the origin. Find the work done in moving the particle from origin to a distance of 9 ft. For this, I set up the integral from 0 to 9 of 10/(1+x)^2 dx and solved, but did not get the final answer of 9ft*lb.
• February 13th 2008, 04:50 AM
topsquark
Quote:

Originally Posted by vperera
A cable that weighs 2ft/lb is used to lift 800lb of coal up a mine shaft 500ft deep. Find the work done. I can't figure out where to start. I understand that W=Fd.

You are lifting 800 lb of coal by 500 ft. Calculating that part of the work is easy.

To get the work done in lifting the cable, note that when you've lifted x ft of cable, you've got 500 - x ft of cable left to lift, which has a weight of 2(500 - x) lb. Thus
$W_{cable} = \int_0^{500}2(500 - x)dx$

-Dan
• February 13th 2008, 04:55 AM
topsquark
Quote:

Originally Posted by vperera
I also have a problem trying to solve something that seems relatively simple: A particle is moved along the x-axis by a force that measures 10/(1+x)^2 lbs at a point x ft. from the origin. Find the work done in moving the particle from origin to a distance of 9 ft. For this, I set up the integral from 0 to 9 of 10/(1+x)^2 dx and solved, but did not get the final answer of 9ft*lb.

$\int_0^9 \frac{10}{(1 + x)^2}~dx = 10 \cdot \int_0^9 \frac{dx}{(1 + x)^2}$

$= 10 \cdot \frac{-1}{1 + x} | _0 ^9$

$= 10 \cdot \left ( \frac{-1}{10} - \frac{-1}{1} \right ) = 10 \cdot \frac{9}{10} = 9$

So I get that the work done is 9 ft-lb.

-Dan
• February 13th 2008, 05:18 AM
vperera
Quote:

Originally Posted by topsquark
You are lifting 800 lb of coal by 500 ft. Calculating that part of the work is easy.

To get the work done in lifting the cable, note that when you've lifted x ft of cable, you've got 500 - x ft of cable left to lift, which has a weight of 2(500 - x) lb. Thus
$W_{cable} = \int_0^{500}2(500 - x)dx$

-Dan

Thanks, but where is the 800lbs accounted for?
• February 13th 2008, 05:19 AM
topsquark
Quote:

Originally Posted by vperera
Thanks, but where is the 800lbs accounted for?

Simply add the 800 * 500 ft-lbs to the amount of work done to lift the cable.

-Dan