# 1D Work Integral

• Mar 8th 2007, 08:09 PM
Aryth
1D Work Integral
Let me see if I have this right: I have to make sure so I don't get everything wrong:

W_AB = ∫_A,^B Fdx

Where F = ma and dx = vdt, and a = dv/dt

W_AB = ∫_A,^B m(dv/dt)*vdt

W_AB = ∫_A,^B mvdv

W_AB = 1/2 mv^2 |_A,^B

W_AB = 1/2 m(v_B^2 -v_A^2)

Yes, no, maybe?
• Mar 9th 2007, 08:13 AM
topsquark
Quote:

Originally Posted by Aryth
Let me see if I have this right: I have to make sure so I don't get everything wrong:

W_AB = ∫_A,^B Fdx

Where F = ma and dx = vdt, and a = dv/dt

W_AB = ∫_A,^B m(dv/dt)*vdt

W_AB = ∫_A,^B mvdv

W_AB = 1/2 mv^2 |_A,^B

W_AB = 1/2 m(v_B^2 -v_A^2)

Yes, no, maybe?

One slight problem: "Fdx" is a dot product between two vectors. This could be negative as well as positive, so you have a second case to consider if you are going to use just the magnitudes. You also have a problem with the integration limits in the last integral. I would recommend changing your proof in the following way:

W_AB = ∫_A,^B F (dot) dx

Where F = ma and dx = vdt, and a = dv/dt

W_AB = ∫_A,^B m(dv/dt) (dot) vdt

W_AB = ∫_v(A),^v(B) mv (dot) dv
where v(A) is the speed at point A, etc. (We need to make this distinction to ensure the correct units at the end.)

Now, either v (dot) dv is positive or it is negative. (If v (dot) dv is always 0 then v and dv are perpendicular and the work done is 0 J.) If v and dv are in the same direction then v(A) < v(B) and the integral is positive. If v and dv are in opposite directions then v(A) > v(B) so the integral has an extra negative sign. (I'll leave you to do the case where the relative sign of v and dv changes over the integration interval.) Either way, the overall integration thus gives a positive value. So we may write:
W_AB = ∫_v(A),^v(B) mvdv

W_AB = 1/2 mv^2 |_v(A),^v(B)

W_AB = 1/2 m(v_B^2 -v_A^2)

-Dan
• Mar 9th 2007, 08:37 AM
Aryth
Quote:

Originally Posted by topsquark
One slight problem: "Fdx" is a dot product between two vectors. This could be negative as well as positive, so you have a second case to consider if you are going to use just the magnitudes. You also have a problem with the integration limits in the last integral. I would recommend changing your proof in the following way:

W_AB = ∫_A,^B F (dot) dx

Where F = ma and dx = vdt, and a = dv/dt

W_AB = ∫_A,^B m(dv/dt) (dot) vdt

W_AB = ∫_v(A),^v(B) mv (dot) dv
where v(A) is the speed at point A, etc. (We need to make this distinction to ensure the correct units at the end.)

Now, either v (dot) dv is positive or it is negative. (If v (dot) dv is always 0 then v and dv are perpendicular and the work done is 0 J.) If v and dv are in the same direction then v(A) < v(B) and the integral is positive. If v and dv are in opposite directions then v(A) > v(B) so the integral has an extra negative sign. (I'll leave you to do the case where the relative sign of v and dv changes over the integration interval.) Either way, the overall integration thus gives a positive value. So we may write:
W_AB = ∫_v(A),^v(B) mvdv

W_AB = 1/2 mv^2 |_v(A),^v(B)

W_AB = 1/2 m(v_B^2 -v_A^2)

-Dan

Thanks a lot. I got that after the second try... Just have to keep trying. It's much clearer now, the only thing is, my teacher left out the fact it was a dot product until we did the later version:

W_(infinity) -> P = ∫_(infinity),^P F(dot)dr