When a particle is located a distance x meters from the origin, a force of cos(PIx/3) Newtons acts on it. How much work is done in moving the particle from x = 1 to x =2? Interpret your answer by considering the work done from x =1 to x = 1.5 and from x = 1.5 to x = 2
So the integral isn't a problem.
If you graph this function I'm sure you know the sign change where cos(pix/3) dips below the x-axis at 1.5. I'm just wondering if I should give my answer as the sum of the neg work and positive work. If the work is negative doesn't it mean it is on the opposite direction? But I still say oh this has done. Neg + pos work