Resource Lesson
APC: Work Notation
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The
work
done on an object by an external force is given by the formula
work
_{done}
= 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
work
_{done}
= ||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
work
_{done}
= F
_{x}
* s
_{x}
+ F
_{y}
* s
_{y}
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:
work
_{done}
= ||F|| * ||s|| cos(θ)
work
_{done}
= F
_{x}
* s
_{x}
+ F
_{y}
* s
_{y}
Method #1
: work
_{done}
= ||F|| * ||s|| cos(θ)
||F|| = 10 N
||s|| = 2 m
θ = 37º + | -23º | = 60º
work
_{done}
= ||F|| * ||s|| cos(θ) = (10)(2) cos (60º) = 10 J
Notice that this result could alternatively be calculated with the expression:
work
_{done}
= [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
: work
_{done}
= F
_{x}
* s
_{x}
+ F
_{y}
* s
_{y}
F
_{x}
= F cos(α) = 10 cos(37º) = 8 N
F
_{y}
= F sin(α) = 10 sin(37º) = 6 N
s
_{x}
= s cos(β) = 2 cos(-23º) = 1
.
84 m
s
_{y}
= s sin(β) = 2 sin(-23º) = -0
.
781 m
work
_{done}
= F
_{x}
* s
_{x}
+ F
_{y}
* s
_{y}
= (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 = 90
^{o}
the work
_{done}
= 0 since cos(90
^{o}
) = 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)
work
_{done}
= Fs
Work can also be calculated as
W
_{done}
=
∫
F ds
or the equivalent area under a force vs displacement graph.
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