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**Work can be either positive or negative: if the force has a component in the same direction as the displacement of the object, the force is doing positive work. If the force has a component in the direction opposite to the displacement, the force does negative work.**

In the light of above statement I have some doubt.

Let's say we've some kind of field which is giving rise to a force ##\mathbf{F} = kx^2 \hat i## (k is some positive constant) on some particle depending on it's position x.

If

*I*move the particle from ## x =0## to ##x=-2## then it is certain that I've to apply a force of magnitude ##kx^2## but opposite in direction of that

*field force*. The work that I'll do will be the negative of the work done by the field force and from the statement that I have quoted we can say that field force will do a negative work because force is in ##\hat i## direction and displacement will be in directly opposite direction, i.e. ##- \hat i##.

Work done by the field force is

$$ W = \int_{0}^{-2} \mathbf{F} \cdot d\mathbf{l} \\

\\

\mathbf{F} \cdot d\mathbf{l} = \langle kx^2 , 0 , 0 \rangle \cdot \langle -dx , 0 , 0 \rangle \\

\textrm{ I've written -dx because displacement is in negative x direction} \\

\\

\mathbf{F} \cdot d\mathbf{l} = -kx^2 ~dx + 0 +0 \\

\\

W = \int_{0}^{-2} -kx^2 ~dx \\

W = -\frac{kx^3}{3} \bigg |_{0}^{-2} \\

W = -\frac{k~(-2)^3}{3} - \left( \frac{-k 0^3}{3}\right) \\

W = \frac{-k ~ (-8)}{3} \\

W = \frac{8k}{3}$$

I have got positive work! WHY?

Thank you. Any help will be much appreciated.