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

  W = Fs

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

        W = Fs = 165(2) = 330 foot pounds.


Hook's Law

For a spring

 F = kd 

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.


 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?


We first find k:

        50 = k(2)  


        k = 25

We have 

        F = 25d  

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

Adding the small intervals, we have that




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?



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

 F = k/x2  

We have that 

        50 tons  =                 

or that 

        k = 800,000,000.

We have that  

        W = Fs 

Cutting the work into small intervals and adding, we have


Emptying a Tank


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


We use the formula        

        F  =  weight  =  (density)(volume) 

Considering cross-sections of the tank, we have

        DF  =  (64 lbs per cu ft)p r2 Dh cu ft  =  64p(100)Dy

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

        (20 - y) feet.

We have



Lifting a Chain



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?  


        Work = Work for the car + Work for the chain

We calculate:

        Work for the car = 2,000(100)

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



Hence the total work is

        200,000 + 50,000 = 250,000 ft lbs.


Running an Automobile


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

        Force = pressure (lbs/sq inch) times Volume ( cu inches) = (pressure)(p r2)

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

        P = k/V


        DW  =            p r2 Dx


        p r2D x = D V

Hence the total work is



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.


We have 

        P = k/v

Plugging in gives

        800 = k/0.5

So that

        k = 400

Now we compute the work:



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