The primary test for determining whether a force is conservative or nonconservative is this: Let the force act on a particle that moves along any closed path, beginning at some initial position and eventually returning to that position (so that the particle makes a round trip beginning and ending at the initial position). The force is conservative only if the total energy it transfers to and from the particle during the round trip along this and any other closed path is zero. In other words:
The net work done by a conservative force on a particle moving around any closed path is zero.
We know from experiment that the gravitational force passes this closed-path test. An example is the tossed tomato of Fig. 8-2. The tomato leaves the launch point with speed v0 and kinetic energy 
. The gravitational force acting on the tomato slows it, stops it, and then causes it to fall back down. When the tomato returns to the launch point, it again has speed v0 and kinetic energy 
. Thus, the gravitational force transfers as much energy from the tomato during the ascent as it transfers to the tomato during the descent back to the launch point. The net work done on the tomato by the gravitational force during the round trip is zero.
An important result of the closed-path test is that:
The work done by a conservative force on a particle moving between two points does not depend on the path taken by the particle.
For example, suppose that a particle moves from point a to point b in Fig. 8-4a along either path 1 or path 2. If only a conservative force acts on the particle, then the work done on the particle is the same along the two paths. In symbols, we can write this result as
Fig. 8-4 (a) As a conservative force acts on it, a particle can move from point a to point b along either path 1 or path 2. (b) The particle moves in a round trip, from point a to point b along path 1 and then back to point a along path 2.
where the subscript ab indicates the initial and final points, respectively, and the subscripts 1 and 2 indicate the path.
This result is powerful because it allows us to simplify difficult problems when only a conservative force is involved. Suppose you need to calculate the work done by a conservative force along a given path between two points, and the calculation is difficult or even impossible without additional information. You can find the work by substituting some other path between those two points for which the calculation is easier and possible. Sample Problem 8-1 gives an example, but first we need to prove Eq. 8-2.
Proof of Equation 8-2
Figure 8-4b shows an arbitrary round trip for a particle that is acted upon by a single force. The particle moves from an initial point a to point b along path 1 and then back to point a along path 2. The force does work on the particle as the particle moves along each path. Without worrying about where positive work is done and where negative work is done, let us just represent the work done from a to b along path 1 as Wab,1 and the work done from b back to a along path 2 as Wba,2. If the force is conservative, then the net work done during the round trip must be zero:
and thus
In words, the work done along the outward path must be the negative of the work done along the path back.
Let us now consider the work Wab,2 done on the particle by the force when the particle moves from a to b along path 2, as indicated in Fig. 8-4a. If the force is conservative, that work is the negative of Wba,2:
Substituting Wab,2 for – Wba,2 in Eq. 8-3, we obtain
which is what we set out to prove.
CHECK POINT 1 The figure shows three paths connecting points a and b. A single force 
does the indicated work on a particle moving along each path in the indicated direction. On the basis of this information, is force conservative?
Figure 8-5a shows a 2.0 kg block of slippery cheese that slides along a frictionless track from point a to point b. The cheese travels through a total distance of 2.0 m along the track, and a net vertical distance of 0.80 m. How much work is done on the cheese by the gravitational force during the slide?
Solution: A Key Idea here is that we cannot use Eq. 7-12 (Wg = mgd cos 
) to calculate the work done by the gravitational force 
as the cheese moves along the track. The reason is that the angle between the directions of and the displacement 
varies along the track in an unknown way. (Even if we did know the shape of the track and could calculate along it, the calculation could be very difficult.)
A second Key Idea is that because is a conservative force, we can find the work by choosing some other path between a and b—one that makes the calculation easy. Let us choose the dashed path in Fig. 8-5b; it consists of two straight segments. Along the horizontal segment, the angle is a constant 90°. Even though we do not know the displacement along that horizontal segment, Eq. 7-12 tells us that the work Wh done there is
Wh = mgd cos 90° = 0.
Along the vertical segment, the displacement d is 0.80 m and, with and both downward, the angle is a constant 0°. Thus, Eq. 7-12 gives us, for the work Wv done along the vertical part of the dashed path,
Fig. 8-5 (a) A block of cheese slides along a frictionless track from point a to point b. (b) Finding the work done on the cheese by the gravitational force is easier along the dashed path than along the actual path taken by the cheese; the result is the same for both paths.
The total work done on the cheese by as the cheese moves from point a to point b along the dashed path is then
This is also the work done as the cheese slides along the track from a to b.
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