PhysicsLAB Resource Lesson
APC: Work Notation

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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)
  • ||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:
  1. workdone = ||F|| * ||s|| cos(θ)
  2. 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 N
sx = 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.

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