Work from a Constant Force

Yes, in this case, if you resolve it parallel and perpendicular to the direction of the force, and only use the component in the direction of the force to calculate the work done.

However, I would recommend not doing this, but stick to resolving the force instead, as it is less likely to lead to confusion in more complicated situations.

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Since the component of force in the direction of movement is now Fcosθ, the work done is just FLcosθ

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But how you determine that the component of force is in the direction of movement  ?  

Resolve the (left-facing, horizontal) force parallel to the line L and perpendicular to it. Only the parallel component counts towards the work done.  The angle θ is the smaller angle between F and L so the component Fcosθ points along L in the direction from A to B.

This is the same direction in which it pointed in part A of the question.  However, the movement is now in the opposite direction from part A, so it's not surprising that the sign is just the opposite of that in part A.

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Thank you alan 

now its clear 

Alan , can I resolve the displacement vector  ? 

well . 

Thank you Alan .