# Force position Graph Area

For question 20 a) it asks to find the work done. I know I have to find the total area under the graph but I'm not sure whether I have to add or subtract the area of the part that is below the x axis.

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• Subtract. The area below the axis is considered work done by the system as opposed to work done to the system. Commented Dec 3, 2016 at 19:47

Graphs were invented to help give us a mental picture the physical actions and relationships that they represent.

Work is the transfer of energy to or from a system or object by a force exerted through a distance.   Work that transfers energy to the system is positive, and work that transfers energy out of it is negative.   So the work done is the positive work minus the negative work (speaking mathematically, that's the positive plus the negative).

When the force is negative, it is actually counteracting the displacement. It is indeed working against it, doing negative work which naturally must be subtracted.

When in doubt, always think of the original reason that you find the area - The reason is the formula for work:

$$W=\int \vec F\cdot d\vec x$$

Integrals are always the area under the graph. If you think of this area as many, many, many thin coloumns packed close, then the formula shows height times width of each coloumn; in this case force times displacement - which is the area of each coloumn. And the integral-symbol $\int$ then sums it all up to the total area.

Now think about it in your case:

• If the force i negative, then you have negative force times displacement. This of course gives a negative result for each of those coloumns. When all is summed together, these therefore subtract from the total area.