Trapezoidal rule to find total work?

I'm given 6 values for time, Force, and velocity. And I'm told to find total work with trapezoidal rule. (first time value is zero)

Does this seem correct? I excluded the code where I assign t=[0,#, #,...] & v=[.2, #, #...] & F=[2.0, #, #...]

pos(1)=0
work(1)=0
totalwork=0
for i=2:length(t)
    area=(v(i)+v(i-1))*(t(i)-t(i-1))/2
    pos(i)=pos(i-1) + area
    work(i)= (pos(i) + pos(i-1))*(F(i)-F(i-1))/2
    totalwork= totalwork + work(i)
end 

totalwork

[David Goodmanson]

Hi Rachel,

you should take a look at the (pos(i) + pos(i-1))*(F(i)-F(i-1))/2 term. If the force is constant everywhere, your expression will produce zero for the total work. So you need to describe the correct trapezoid.

[Rachel Dawn]

Hi David, thank you for your response. The Force values I'm given are not constant throughout. But, I'm wondering, how would I actually change my code to account for a constant force? What do you mean by "describe the correct trapezoid"?

Thanks!

Answer 1

[Roger Stafford]

I would think your code should be this:

work = 0;
for k = 2:length(t)
  work = work + (F(k)+F(k-1))/2*(v(k-1)+v(k))/2*(t(k)-t(k-1));
end

That is, the quantity "(v(k-1)+v(k))/2*(t(k)-t(k-1))" is the approximate displacement during the time interval t(k-1) to t(k), and if it is multiplied by the average force (trapezoid rule), "(F(k)+F(k-1))/2", during that time interval you would get the approximate work done then. The sum of the five work values should give you the total work done.

[Addendum: Or perhaps you could use this:

work = 0;
for k = 2:length(t)
  work = work + (F(k)*v(k)+F(k-1)*v(k-1))/2*(t(k)-t(k-1));
end

because you are approximating the integral of F*v with respect to time, t.]

[Rachel Dawn]

Hi Roger, Thank you! I checked this code and got a totally different answer for work then my previous code. What was wrong with mine, though?

Hmm. I just tried both those sections of code you included and they give different answers. I'm not sure why.

Update: I figured out what was wrong with my code! I had position as my bases & force as my height (should be the other way around). Thanks!

[Roger Stafford]

" I just tried both those sections of code you included and they give different answers." Yes, they are not identical, but are different approximations. Assuming F and v vary in a reasonably smooth fashion, they should not be greatly different. It is the difference between

(F(k)*v(k)+F(k-1)*v(k-1))/2

and

(F(k)*v(k)+F(k)*v(k-1)+F(k-1)*v(k)+F(k-1)*v(k-1))/4

It is not clear which of these best represents the trapezoidal rule. I would hazard the guess that the first of these (that is, the second in the answer) is likely to be the best.