┴
F
g ||
θ
F = F cosθ =
g|
g
g
m cosθ
F
g = mg
F
= F sinθ =
g⊥
g
g
m sinθ
|F |
g ⊥
||
Now we redraw the sketch with the gravitational force replaced by its components:
A
θ
d
F
g ||
|F |
g ⊥
B
Fg , being perpendicular to the path does no work on the block as the block moves from A to B.
⊥
The work done by the gravitational force is given by
W = F d
|
W = F d
g|
W = mg(cosθ ) d
W = mgd cosθ
While this method for calculating the work done by a force is perfectly valid, there is an easier
way. It involves another product operator for vectors (besides the cross product), called the dot
product. To use it, we need to recognize that the length of the path, combined with the direction
of motion, is none other than the displacement vector (for the point of application of the force).
Then we just need to find the dot product of the force vector and the displacement vector.
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