How do you calculate the work done on an interval when given a function?

Question

Say you had: 4x³-28x and the interval x=1m to x=2m how would you find work?

Answer 1

Hi Katlyn K.,

"Work" done on an interval only makes sense if you know the force applied over the interval! Otherwise, you could do the same integration of a function, but the result is just called an integral -- it has a mathematical meaning (and a graphical meaning (it's the area between the vertical lines of the interval, the x-axis, and the function curve)), but not necessarily a physics meaning.

Let's assume that the function supplied represents the force, and that it is applied in the positive x-direction (it wouldn't necessarily have to be, in a physics problem!). So technically:

F = (4x^3 - 28x)x

where F has a vector arrow over it, and the x at the right has a hat (^) to indicate that it's the unit vector in the x-direction. Sorry I don't know the various scripts well enough to write that ...

Then you obtain the first integral of the function expression, and evaluate that new function over the interval, by subtracting the value at the start of the interval from the value at the end of the interval.

So: integral of (4x^3 - 28x)dx is (remember, integral of each term ax^n is a*x^(n+1)/(n+1):

4*x^4/4 - 28x^2/2 = x^4 - 14x^2

Evaluate over the interval:= (2^4 - 14*2^2) - (1^4 -14*1^2) = (16 - 56) - (-13) = -27 Q.E.D. -- But don't forget your pseudo-work units, m^2/s, since the mass was not specified!

Remember, always inspect your problem statements for any possible hidden assumptions -- probably at your course level you would be expected to make those assumptions without being aware of alternatives, but at higher course levels you will be required to state all assumptions made, and possibly probe alternatives to those assumptions. Example: solutions to x^2+4 = 0 . At beginning algebra levels, you would say that this has no solutions, but at algebra 2 levels you would be expected to recognize and utilize the complex numbers to solve to ± 2i .

-- Cheers, -- Mr. d.

Answer 2

If you are given a function F(x) and are asked to find the work over an integral, you need to integrate the function. The work is just the integral from a to b of F(x)dx. So here we would have the integral from 1 to 2 of 4x^3-28x. So we just integrate, plug in the limits and solve.

Note: these problems get hard when they DONT give you the function and you have to interpret it from the graph or some other features of the problem.