# Why is the work positive here?

Problem

http://apcentral.collegeboard.com/apc/public/repository/ap11_frq_physics_b_formb.pdf

Please refer to question 1f Solutions http://apcentral.collegeboard.com/apc/public/repository/ap11_frq_physics_b_formb.pdf

Question from me

Isn't he climbing up? Gravity points down and he is going up, shouldn't it be -mgh instead of +mgh? He is doing work AGAINST gravity isn't he?

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If this is a homework question, please add the "homework" tag. And please type the problem in full so we don't have to check an eleven-page document. Plus, you got wrong the link for the solutions. – Arnoques Dec 12 '11 at 22:19
The sign of work is obvious from intuition. Anytime A does work on B, B does negative work on A (since energy is conserved). Don't get anxious over signs here. – Ron Maimon Dec 13 '11 at 14:36

You are correct. However you are ignoring the sign associated with the variable $g$. If gravity points down you need another negative sign, which results in positive work done.
For your problem, notice that it asks you for the work that the person does on the object. For a constant force and a rectilinear movement, the work is $L=F\,d\,\cos(\theta)$, where $F$ is the force in question, $d$ is the distance traveled and $\theta$ is the angle between the force and the direction of displacement. Notice that both $F$ and $d$ are positive in the equation, so the sign comes from the $\cos(\theta)$. If the force points to the direction of movement (i.e., if it "helps" the movement) the work will be positive. If the force points backwards (i.e., it "opposes" the movement) the work will be negative.