# Calculating Work Done by Variable Force

1. 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.

2. Homework Equations

see above, and maybe w=fd

3. 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?
1. Homework Statement

2. Homework Equations

3. The Attempt at a Solution

## Answers and Replies

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berkeman
Mentor
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.

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 beleive thats 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. Im wanting to know if that answer is the work done or not.

berkeman
Mentor
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.

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.