Understanding Work in Physics
Physicists Multiply Force by Displacement
In physics work has a very specific definition that is different from the everyday definition. A force must be applied over a distance.
Work in Physics
Who does more work; a weightlifter holding a 150 pound weight overhead perfectly motionless for an hour, or someone lifting a flea one centimeter?
Nonphysicists are likely to say that the weightlifter does more work. The weightlifter certainly exerts more effort. However a weightlifter holding the weight perfectly motionless is, according to the physics definition of work, doing no work. Someone lifting a flea one centimeter does a small, but nonzero, amount of work.
The key to understanding this apparent paradox is knowing that to physicists work has a meaning different from the everyday definition. Physicists find work by multiplying the force times the distance over which the force is applied.
The weightlifter is exerting a large force, but the distance is zero. Hence the weightlifter does zero work, despite the obvious effort and exertion. The flea lifter exerts a small force over a small distance. Multiplying these two numbers gives a small but nonzero work. So lifting a flea a small distance is more work that holding a heavy weight stationary.
Mathematical Formulas for Work
The simplest mathematical formula for computing the amount of work arises when the applied force, F, is constant and in the same direction as the displacement, d. In this case find the work, W, that the force does using the formula:
W = F d
If sliding a large box 3 meters across a floor requires a 10 newton horizontal force, then the work is:
W = (10 newtons)X(3meters) = 30 newton meters = 30 joules
When the applied force and the displacement vector are in different directions, the work can be found using the equation:
W = F d cos(theta)
Theta is the angle between the force vector and the displacement vector. In more mathematical terminology, the work is the dot (or scalar) product of the force and displacement vectors.
Notice that if the force and displacement vectors point in opposite directions, then theta is 180 degrees and cos theta is -1. In this case the force does a negative work. If one is lowering a heavy box to the floor, then one might be applying a 50 newton upward force while the box moves 1 meter downward. The work this force does on the box is given by:
W = F d cos(theta) = (50 newtons)X(1 meter) cos(180 degrees) = -50 joules
Net Work
If there is more than one force acting on an object the net (or total) work done on the object is found by using the vector sum of all the forces acting on the object in the above formula. The net work done on an object is also equal to the change in the object's kinetic energy.
For example when lowering the 50 newton box, at a constant speed, in the above example, Earth's gravity pulls down on the box with a 50 newton force. The person lowering the box applies a 50 newton upward force. The net force and therefore net work on the box are both zero. Earth's gravity does 50 joules of work on the box which when added to the -50 joules done by the person gives a net work of 0 joules.
Variable Force
The force isn't always constant. In the case of a variable force, one must use calculus techniques to find the work. Integrate F dx over the entire displacement to find the work.
Remember that in physics work has a very specific meaning different than the everyday definition.
Further Reading
Knight, R.D., Physics for Scientists and Engineers with Modern Physics, Pearson, 2004.
Understanding Potential Energy
