the object’s displacement vector, then F| is negative, so the work is negative. Thus, if you are
pushing or pulling on an object in a direction that would tend to make it speed up, you are doing
positive work on the object. But if you are pushing or pulling on an object in a direction that
would tend to slow it down, you are doing negative work on the object.
In the most general case in which the “component of the force along the path” is continually
changing because the force is continually changing (such as in the case of an object on the end of
a spring) or because the path is not straight, our “how-to-calculate-it” definition of the work
becomes: For each infinitesimal path segment making up the path in question, we take the
product of the along-the-path force component and the infinitesimal length of the path segment.
The work is the sum of all such products. Such a sum would have an infinite number of terms.
We refer to such a sum as an integral.
The Relation Between Work and Motion
Let’s go back to the simplest case, the case in which a force F is the only force acting on a
particle of mass m which moves a distance ∆r (while the force is acting on it) in a straight line in
the exact same direction as the force. The plan here is to investigate the connection between the
work on the particle and the motion of the particle. We’ll start with Newton’s 2nd Law.
Free Body Diagram
F
m
a
1
a
=
F
→
∑ →
m
1
a =
F
m
Solving for F, we arrive at:
F = m a
On the left, we have the magnitude of the force. If we multiply that by the distance ∆r , we get
the work done by the force on the particle as it moves the distance ∆r along the path, in the same
direction as the force. If we multiply the left side of the equation by ∆r then we have to multiply
the right by the same thing to maintain the equality.
F ∆r = m a ∆r
1
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