5.2: Work

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Definition of Work

Definition: Work

We define the work, W, done in moving an object a distance s units with a force of $F$ as:

\[W = Fs.\nonumber \]

Example 1

For me to jump 2 feet in the air, if I weigh 165 pounds, takes

\[W = Fs = 165(2) = 330 \text{ foot pounds}.\nonumber \]

Hooke's Law

For a spring

\[F = kd\nonumber \]

where $d$ is the distance the spring is from its equilibrium point, $F$ is the force exerted to compress to that distance, and $k$ is a proportionality constant.

Example 2

To compress a spring 2 inches from its equilibrium point requires 50 pounds of force. How much work is it to compress the spring an additional 2 inches?

$work1.13.gif$

Solution

We first find $k$:

\[50 = k(2)\nonumber \]

hence

\[k = 25\nonumber \]

We have

\[F = 25d.\nonumber \]

To find the work done, we cut the work into small intervals of work. The force required to move the spring $Dx$ units is $25x$, hence the work required is

\[Dw = 25x \; Dx.\nonumber \]

Adding the small intervals, we have that

\[\begin{align*} W &= \int_2^4 25x \; dx \\ &= \left(\dfrac{25}{2}x^2\right]_2^4 \\ &= 200-50 \\ &= 150. \end{align*} \nonumber \]

Rockets

Example 3: Sending a Rocket to Mars

Suppose that a 50 ton rocket is to be sent to mars and after 100,000 miles, the rocket is completely free of the earth's gravity. If the radius of the earth is 4,000 miles, how much work is required to send the rocket out of orbit?

$rocket.jpg$

Solution

We use that the force of gravity is inversely proportional to the square of the distance from the center of the earth.

\[F=\dfrac{k}{x^2} \nonumber \]

We have that

\[50 \text{tons} = \dfrac{k}{4000^2} \nonumber \]

or that

\[ k = 800,000,000.\nonumber \]

We have that

\[W = Fs.\nonumber \]

Cutting the work into small intervals and adding, we have

\[\begin{align*} W&= \int F \; dx \\ &= \int _{4,000}^{100,000} \dfrac{80,000,000}{x^2} \; dx \\ &= 192,307 \text{ mile-tons}. \end{align*} \nonumber \]

Emptying a Tank

Example 4

Suppose that a cylindrical tank of water with radius 10 and height 20 is pumped out fully. How much work is required?

Solution:

We use the formula

\[F = weight = (density)(volume).\nonumber \]

Considering cross-sections of the tank, we have

\[\begin{align*} DF &= (64 \text{ lbs per cubic ft})pr^2Dh \text{ cubic ft} \\ &= 64p(100) \;Dy. \end{align*} \nonumber \]

The distance that water $y$ feet above the ground must be pumped is

\[(20 - y) \text{ feet}.\nonumber \]

We have

\[\begin{align*} W&= \int_0^{20} 6400 \pi (20-y) dy \\ &= 6400 \pi \left(20y-\dfrac{1}{2}y^2 \right]_0^{20} \\ &= 6400 \pi (400-200) \\ &= 1,280,000 \text{ foot-pounts}. \end{align*} \nonumber \]

Lifting a Chain

Example 5

A 2,000 lb car has gone over a 100 foot cliff at the pass. A crane lifts up to car with a chain that is 10 lbs per foot. How much work is done?

Solution:

\[\text{Work } = \text{ Work for the car} + \text{ Work for the chain} \nonumber \]

We calculate:

\[\text{Work for the car} = 2,000(100)\nonumber \]

and since the force is 10 times the distance the chain link is from the top

\[\begin{align*} \text{Work for the chain} &= \int_0^{100} 10x \; dx \\ &= \left( 5x^2 \right] \\ &= 50,000. \end{align*} \nonumber \]

Hence the total work is

\[ 200,000 + 50,000 = 250,000 \text{ft lbs}. \nonumber \]

Running an Automobile

An automobile is run by a piston which turns the wheel. We have the equation

\[Force = pressure (\text{lbs/sq inch}) times Volume ( \text{cu inches}) = (pressure)(p r^2). \nonumber \]

Boyle's Law states that the pressure is inversely proportional to the volume:

\[P= \dfrac{k}{V}. \nonumber \]

Hence

\[DW=\dfrac{k}{V}pr^2 \; Dx .\nonumber \]

But

\[pr^2 \; Dx = DV. \nonumber \]

Hence the total work is

\[W= \int_a^b \dfrac{k}{V} \; dV. \nonumber \]

Example 6

Suppose a piston has gas filled initially to a volume of 0.5 cu ft and a pressure of 800 lbs/sq ft. The piston expands to a volume of 1 cu ft. Find the work done by the gas.

Solution

We have

\[P = \dfrac{k}{v}. \nonumber \]

Plugging in gives

\[800 = \dfrac{k}{0.5}. \nonumber \]

So that

\[k = 400.\nonumber \]

Now we compute the work:

\[ \begin{align*} W&= \int_{0.5}^{1} \dfrac{400}{V}\; dV \\ &= \left( 400 \ln V \right]_{0.5}^1 \\ &= 400 \ln 1 - 400 \ln 0.5 \\ &= 277 \text{ft lbs.} \end{align*} \nonumber \]

Larry Green (Lake Tahoe Community College)

Integrated by Justin Marshall.