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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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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. |
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I disagree with Ron Maimon's comment. The sign of the work is important because it tells you who is gaining energy, and who is losing it. 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. Gravity does work on the object also, but that has nothing to do, in principle, with the work of this force. To see the relationship between both works, you need to check what happens with the mechanical energy of the box before and after it moves. |
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