Work

The work required to raise a weight of P pounds a distance of y feet is $P\cdot y$ foot-pounds. (In m-k-s units, one would say that a force of k newtons exerted over a distance of y feet does $k\cdot y$ newton-meters, or joules, of work.)

Example. If a 100 pound weight is lifted a distance of 3 feet, the work done is 300 foot-pounds.

Example. A 6-foot cable weighing 0.6 pounds per foot is suspended by one end from a reel. How much work is done in winding the entire cable onto the reel?

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Consider a small piece of the cable of length which is y units above the bottom end (before the cable is wound up). The weight of the piece is $0.6\,dy$ , and it must be raised a distance of feet to get to the reel. The work done in lifting this piece is $(6 - y)(0.6\,dy)$ .

The total work required is

$\int_0^6 (6 - y)(0.6\,dy) = 0.6\left[6y - \dfrac{1}{2}y^2\right]_0^6 = 10.8\ \hbox{foot-pounds}.\quad\halmos$

Example. Find the work done in pumping all the water out of the top of a filled cylindrical tank of radius 3 feet and height 6 feet.

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The picture on the left shows a side view of the cylinder. I'll cut the cylinder of water up into circular slices; a typical slice is shown on the right.

I'll make the simplifying assumption that to pump the water in the slice out of the tank, I just need to raise the whole thing vertically to the top of the tank.

Water has a density of 62.4 pounds per cubic foot. A typical circular slice of the water weighs

$62.4\cdot \pi\cdot 3^2\cdot dy = 9\pi(62.4)\,dy \quad\hbox{pounds}.$

It is raised a distance of feet.

The work done in raising such a slice is $9\pi(62.4)(6 - y)\,dy$ foot-pounds. Thus, the total work done in emptying the tank is

$W = \int_0^6 9\pi(62.4)(6 - y)\,dy = 162\pi\cdot 62.4 \approx 31757.73182\ \hbox{foot-pounds}.\quad\halmos$

Example. Find the work done in pumping all the water out of a filled sphere of radius 2 feet. Assume that the water is pumped out of the top of the sphere.

$\hbox{\epsfysize=1.75in \epsffile{work3.eps}}$

The picture on the left shows a side view of the sphere. I'll cut the cylinder of water up into circular slices; a typical slice is shown on the right.

In the picture, the slice has radius $r = \sqrt{4 - y^2}$ . The weight of the slice is

$62.4\cdot \pi\cdot (\sqrt{4 - y^2})^2\,dy = 62.4\pi(4 - y^2)\,dy.$

The slice is raised a distance of feet, so the work done is

$62.4\pi(4 - y^2)(2 - y)\,dy.$

The total work done is

$W = \int_{-2}^2 62.4\pi(4 - y^2)(2 - y)\,dy = 62.4\pi\cdot \dfrac{64}{3} \approx 4182.08814\ \hbox{foot-pounds}.\quad\halmos$

Consider a spring attached at one end to a stationary object, such as a wall. Stretch the spring so that its free end moves a distance x from its unstretched position. The force exerted by the spring is given by Hooke's law:

where k is the spring constant. (The negative sign reflects the fact that the force is a restoring force: It always points toward the origin.)

To stretch the spring, I must exert a force equal to and opposite in direction to the force exerted by the spring: . Suppose I do so, and the end of the spring is moved by a small amount . Then the work done is $F\,dx = kx\,dx$ . Hence, the total work done in stretching the spring from to is

$W = \int_a^b kx\,dx = \left[\dfrac{1}{2}kx^2\right]_a^b = \dfrac{1}{2}ka^2 - \dfrac{1}{2}kb^2.$

Example. A force of 12 pounds is required to stretch a spring 1.5 feet beyond its unstretched length.

(a) What is the spring constant for this spring?

Using , I have

$12 = 1.5k, \quad k = \dfrac{12}{1.5} = 8\ \hbox{pounds per foot}. \quad\halmos$

(b) How much work is done in stretching the spring from 2 feet beyond its unstretched length to 4 feet beyond its unstretched length?

$W = \int_2^4 8x\,dx = \left[4x^2\right]_2^4 = 64 - 16 = 48\ \hbox{foot-pounds}.\quad\halmos$

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Last updated: December 2, 2005

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