Calculating Work Done by Variable Force

[rock4christ]

Homework Statement

I need To find the work done by a force that varies with position like this:
F(x)=2x^3 -5x^2 +10

with x at values between 0 to 10 meters.

Homework Equations

see above, and maybe w=fd

The Attempt at a Solution

I found the area underneath the curve by putting the values from 0.001 to 10 into Excel, and putting in the equation for all values. it looks like this:

x (in meters) f(x) area of each increment
0.1 9.952 0.009952
0.2 9.816 0.009816
0.3 9.604 0.009604
0.4 9.328 0.009328
0.5 9 0.009
0.6 8.632 0.008632
0.7 8.236 0.008236
0.8 7.824 0.007824
0.9 7.408 0.007408
1 7 0.007
1.1 6.612 0.006612
1.2 6.256 0.006256


area of all increments(using 0.001, not 0.1)
3434.083375

^thats just a sample and is only using .1 increments.

would 3434.083375 actually be my final answer for the work performed, or would that be the force, and if so, what distance would I multiply it by?

 

[berkeman]

If you're going to solve it (approximately) nunerically instead with integration, then for each delta-X, you multiply the force in that delta-X by the delta-X, and add up all those pieces of the work to get the total.

 

[rock4christ]

If you're going to solve it (approximately) nunerically instead with integration, then for each delta-X, you multiply the force in that delta-X by the delta-X, and add up all those pieces of the work to get the total.

I believe that's what I did, because I took my force, of say 6.612, and * it by .001 and did that for all of them, and added it up. I am wanting to know if that answer is the work done or not.

 

[berkeman]

In your relevant equations, you listed W=fd. So it would seem you have calculated the correct thing.

There is a subtlety, though in doing numerical calculations like this. You will get small errors for each delta-X section, because the delta-Work that you are calculating is not exact. That's why you will later learn to use integration to calculate the exact total work -- integration basically uses infinitesimally small delta-X sections (called "dx"). One improvement you could make to your numerical calculation is to use the y value for the middle of each delta-X section, rather than the value at the end of the delta-X section as you are doing now.

 

[rock4christ]

Yeah I'm not in Calculus yet, so Integrals wouldn't be an option even IF he hadn't specifically said we couldn't use it.