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# Work as area under curve

David shows how the area under a force vs. position graph equals the work done by the force and solves some sample problems. Created by David SantoPietro.

## Want to join the conversation?

• So if the line was a curve, than the work done would be the integral of the formula of the line, right?
• It's always equal to the integral, but sometimes the integral is really simple, and you can just use common area formulas.
• But What if the graph was a quarter of a circle, how do you find the work done in that case?
• Ah, true, the question is not primarily focusing the shape of the graph, just the area.

However, Howeveeeeeer, there is a reason why the shape of the graph is chosen in the form of rectangles/triangles, because they're easy to work with, the equations which describe them are simple, relatively.

So is the equation for a circle (thus any fraction of said circle), but where this breaks down for us, is where the equation for the graph is in a shape that we cannot easily calculate via conventional means.

There is a whole system in mathematics dedicated to just this, just this one feature of graphs, it's so important, an entire system has been based around it, which you will need to learn at some point if you delve far enough into physics. This system is called calculus, which is split into two parts, differential calculus, and integral calculus. Sal Khan has done his thing, and has made video series on calculus. You need not delve into calculus right now, though I would recommend watching at least the first section of differential calculus, as it explains why it was invented. It's origins has to do with the motion of objects, and was invented by Newton to explain said motion.

By the way, you'll need to know differential calculus to understand integral calculus. So start there :)
• @ he takes away the negative work of pushing the burger to the left then to the right from the total force. However in reality work is work, right or wrong? it still takes extra energy (fuel) to push something back in the same direction you came from, you don't get a credit upon return. Or am I missing the point or taking the term "work" too literally?
• If I understood it right, work = energy put into the system/object.

For the first meter, we apply a negative force in the oposite direction of motion and that means slowing the burger down and this equals removing of some of the kinetic energy from the burger, thus negative work for the first meter.

After that 1 meter we start applying a positive force in direction of motion and that means accelerating the object, increasing its kinetic energy and thus having positive work done on that object.

So negative work means we steal some kinetic energy from the moving burger.
• If you push something to the left, does that make it negative work?
• If I may add a contribution:

The thing about work is, its a scalar. not a vector. So, the direction of work does not make it posisitve or negative

The positive and negative thing about work is more about WHO does the work. For example. If You do work on something, then the work you have done on that thing is positive. If the thing does work on you, then the work that YOU have done on that thing is negative.

Hope that makes sense

IM
• What if the graph was a changing curve with different slopes at different times?
• You would integrate the curve:
`W = ∫F.dx from x1 to x2.`
• To find the area under the graph wouldn't you still need to know calculus depending on the degree of the change in the force?
• For simple graphs you only need basic geometry , but if equation of force is not a regular shape ( say a parabole) , you will need to use integral calculus.
• How would you determine work done by a varying force by using calculus??
• You would take the integral of the force vs. the distance:
W = ∫F.dx where W is the work, F is the force in direction of motion, and x is the total displacement.
• At , Why is the work being done to the burger negative? If the force acting on the burger is going in the same direction as the displacement of the burger, shouldn't the work be positive?
• The X-axis is the position of the hamburger. The graph starts out below the X-axis with a negative value meaning it was initially moving in the opposite direction. therefor, causing negative displacement and ultimately negative work. To find the net work you must add the two together.