 APC: Work Notation Printer Friendly Version
The work done on an object by an external force is given by the formula

workdone = force displacement

Work is our first example of a scalar product or dot product. A dot product occurs when two vectors are multiplied together in such a way as to produce a scalar value. Technically, the above definition for work can be calculated with the equation

workdone = ||F|| * ||s|| cos(θ)

where

• ||F|| represents the magnitude or the length of vector F
• ||s|| represents the magnitude or the length of vector s
• θ represents the magnitude of the angle between F and s.

Or, it can equivalently be evaluated with the formula

workdone = Fx * sx + Fy * sy

where the x- and y-components of F and s are multiplied and then added together.

Let's use an example to show how these two expressions are equivalent.

 Suppose a toy cart is sliding between two rails along the surface of a table while being pulled diagonally by a force F. To determine the work done on the cart by the force, we can use either of these two methods:  workdone = ||F|| * ||s|| cos(θ) workdone = Fx * sx + Fy * sy Method #1: workdone = ||F|| * ||s|| cos(θ)   ||F|| = 10 N||s||  = 2 m θ = 37º + | -23º | = 60º  workdone = ||F|| * ||s|| cos(θ) = (10)(2) cos (60º) = 10 J  Notice that this result could alternatively be calculated with the expression:   workdone = [F cos(θ)] s = [10 cos(60º)](2) = 10 J where F cos(θ) is the component of F in the direction of s.   Method #2: workdone = Fx * sx + Fy * sy  Fx = F cos(α) = 10 cos(37º) = 8 N Fy = F sin(α) = 10 sin(37º) = 6 Nsx = s cos(β) = 2 cos(-23º) = 1.84 m sy = s sin(β) = 2 sin(-23º) = -0.781 m   workdone = Fx * sx + Fy * sy = (8)(1.84) + (6)(-0.781) = 10 J  Any slight numerical differences between the two calculations would be the result of rounding decimal expressions.

The formula that is usually used to calculate the amount of work done on a mass M by a constant force F acting at an angle θ to its displacement s is W = Fs cos(θ) Using this formula when θ = 90o the workdone = 0 since cos(90o) = 0. Referring to the above diagram, the vertical component, F sin(θ), of F would not result in any work being done on mass M since it acts at right angles to the displacement s

Work done on an object by a force acting parallel to its displacement can be expressed as simply (where θ = 0º)

workdone = Fs

Work can also be calculated as Wdone = F ds or the equivalent area under a force vs displacement graph. Related Documents