# How to calculate Work as area under the curve

• loganblacke

## Homework Statement

A force acts on an object of mass .425kg. The force varies with position as shown in the graph that follows.. (Force Y-axis, Position x-axis). Find the work done by the force in moving the object from .40m to 1.20m. The two points are (.40m, 1.6N) and (1.20m, 2.2N)

## Homework Equations

Integral of f(x) dx from a to b = work

## The Attempt at a Solution

How do I find the equation of the curve from the two points to integrate it?? Is the mass relevant?

Usually when you are asked to find the work and you are only given the graph (without any additional information about the force as a function of position like: it is quadratic) they usually mean you can do the "integration" geometrically.
I.e. divide the graph into geometrical shapes (rectangles, triangles) of which you can determine the surface, estimate using the grid by counting squares, etc.

Usually when you are asked to find the work and you are only given the graph (without any additional information about the force as a function of position like: it is quadratic) they usually mean you can do the "integration" geometrically.
I.e. divide the graph into geometrical shapes (rectangles, triangles) of which you can determine the surface, estimate using the grid by counting squares, etc.

I am aware of this method however the graph in this question is a curve, the shape of the curve is similar to the graph of y=Sqrt(x).

You can still use the geometrical method for such a curve. The answer you will get will be an approximation, with the accuracy depending on how well you approximate the curve as made up of little straight segments. But it would still work.

the fact that a work integral is equivalent to an area integral translates in other work problems into a volume integral. then finally one notices that a work integral for say pumping water in a tank up to the top, must be a 4 dimensional volume! this is really cool, and allows one to use this technique to compute the volume of a 4 dimensional sphere.

You can still use the geometrical method for such a curve. The answer you will get will be an approximation, with the accuracy depending on how well you approximate the curve as made up of little straight segments. But it would still work.

Thank you for the help, I just used midpoint rule and approximated and it ended up correct.

the fact that a work integral is equivalent to an area integral translates in other work problems into a volume integral. then finally one notices that a work integral for say pumping water in a tank up to the top, must be a 4 dimensional volume! this is really cool, and allows one to use this technique to compute the volume of a 4 dimensional sphere.

I read your post 3 times and still don't have a clue what you are talking about. We have done several "pumping water" problems but I've never heard of them being 4 dimensional.. Doesn't that deal with shadows??

imagine a plane region below the x axis, and the work integral for pumping the contents, with constant density, of this plane region, up to the x axis. If you look at it, it should be identical, except for a multiple of 2pi, to the integral for computing the volume generated by =revolving this area around the x axis.

Similarly, the work integral for pumping a solid region below the x,y plane up to that plane, is a constant times the volume integral for the 4 dimensional volume generated by revolving the solid region below the x,y plane, around that plane in 4 space.

E.g. one obtains the 3 ball by revolving the bottom half of the unit circle around the x axis. Also one obtains the 4 ball by revolving the bottom half of the 3 ball around the x,y plane (in 4 space).

I think I am remembering this correctly. I noticed it last time i taught the course. This is a nice computation, because it makes computing the volume of the 4 ball easier than doing it by the method of slicing.

you can test it on computing the volume of the 3 ball.