Recall that work is the amount of energy required to perform some action. When the amount of force is constant, work is simply
For example, if a book weighing 22 Newtons (about 5 pounds) is lifted 2 meters, the total work done is (J is the Joule, which equals one Newton-meter).
Consider a situation where the force is not constant. For instance, if one were to lift a weight using a non-negligible rope, there is less rope being pulled up (and hence less force) as the weight goes further up. It is in situations like these that we need a better formula to compute work.
Computing work when the force is not constant requires integration. As in previous sections, the first step is to determine the work element , and then integrate:
Because work arises in a variety of situations, there is not one simple formula for the work element. For different applications the work element will look different. In some situations, it is best to consider a small movement , where the force can be thought of as constant for that small movement, which allows the work element to be expressed as .
The force required to displace a spring varies with the displacement. The further the spring is stretched, the more resistant it becomes to being stretched further. Consider three types of springs:
- Linear. A spring is linear if the force of resistance grows linearly with the displacement. That is,
for some constant , which represents the stiffness of the spring.
- Hard. A spring is hard if the force of resistance grows faster than linearly with the displacement:
- Soft. A spring is soft if the force of resistance grows slower than linearly with the displacement:
Consider for any of these springs what the work element is. When the spring is stretched to , the force of resistance is . For the next infinitesimal amount of stretching , the force can be presumed to be constant:
Therefore, the work element (i.e. the amount of work to stretch the spring the additional amount ) is
Compute the amount of work it takes to stretch a linear spring from rest (when ) to .
Consider a nonlinear, soft spring which exerts a force of Newtons when the spring is stretched to meters. Determine how much work is required to stretch the spring from 1 meter to 3 meters.
Pulling up a rope
In some situations, one must do a little work to determine what is, and then one can integrate, as in the above examples.
Consider a rope which is 100 feet long and density 1 pound/foot. It hangs from a wall which is 50 feet high (so 50 feet of rope runs down the length of the wall and the remaining 50 feet is coiled at the bottom of the wall). How much work (in foot-pounds) is required to pull the rope to the top of the wall?
Work element by slices
In other situations, such as pumping liquid, digging a hole, or piling gravel, a fruitful method for determining the work involved is to consider a slice of the material which is being moved. Determining the weight of the slice, and multiplying by the distance the slice has to be lifted gives the amount of work required for that slice. That is precisely the work element. Integrating over all the slices in the object gives the total amount of work to move that object.
Example: pumping liquid
Consider an inverted conical tank (so the tip of the cone points downward) with base radius 5 feet and height 10 feet. Water is pumped into the tank through a valve at the tip of the cone:
How much work is required to fill the tank with water? Leave the weight density of water as the constant .
Example: digging a hole
Consider two workers digging a hole. How deep should the first worker dig so that each does the same amount of work? Let the weight density of the dirt be the constant , the depth of the hole is , and the cross-sectional area of the hole is the constant (so we assume that the hole does not get any wider or narrower as the workers dig).
Example: gravel pyramid
Compute the amount of work required to build a pyramid of gravel. Assume the gravel is infinitesimal with weight density , and that the pyramid has a square base of side length , and height :
Example: rope revisited
Consider the rope example from above, but this time suppose total feet of rope are hanging from a foot building, where , and let be the weight density of the rope. Compute the work required to lift the rope to the top of the building.
For a different perspective, this time, use a work element which equals the amount of work required to lift an infinitesimal length of rope to the top of the building (this will depend on whether the infinitesimal length of rope is hanging at the beginning or is part of the coil at the bottom of the building). Then integrate along the entire length of rope.
- Consider a conical tank of height 10m. The vertex of the cone is at the bottom, and the base of cone (which is at height 10m) has radius 2m. Let denote the weight density of water. The water inside the tank has height 4m. How much work would it take to pull all the water to the top of the tank?