What mistake am I making when calculating work done by a force?

[songoku]

Homework Statement

Draw the graph of $F=-2x$ and calculate the work done from $[-1,1]$

Relevant Equations

Work = area under the graph

The answer key is zero because the areas are above and below x-axis and have equal magnitude so canceling out each other.

But I am confused about the solution

Area 1 is above x-axis but I think the work done is negative since the sign of $F$ and $x$ is opposite. Work done on area 2 is also negative for the same reasoning so my answer would be -2 J

Where is my mistake? Thanks

 

[SammyS]

Homework Statement:: Draw the graph of $F=-2x$ and calculate the work done from $[-1,1]$
Relevant Equations:: Work = area under the graph

The answer key is zero because the areas are above and below x-axis and have equal magnitude so canceling out each other.

But I am confused about the solution
View attachment 297599

Area 1 is above x-axis but I think the work done is negative since the sign of $F$ and $x$ is opposite. Work done on area 2 is also negative for the same reasoning so my answer would be -2 J

Where is my mistake? Thanks

Ask yourself, "What is the definition of Work?"

 

[songoku]

Ask yourself, "What is the definition of Work?"

Work is the product of displacement and force which is parallel to the displacement

So based on that definition, I think when the force is positive and displacement is negative, the work done will be negative.

Thanks

 

[SammyS]

Work is the product of displacement and force which is parallel to the displacement

So based on that definition, I think when the force is positive and displacement is negative, the work done will be negative.

Thanks

Right. Looks like you've got it.

So, in going from $x=-1$ to $x=0$, for instance, the displacement is positive, even though $x$ is negative.

 

[hutchphd]

For a non constant force the work in going from1 to 2 is $$work=\int _1^2 F(x) \,
dx$$ The increment dx is positive in the +x direction. So the net result result
$$work=\int _{x=-1}^{x=+1} F(x) \,
dx=0$$
is zero

 

[songoku]

Right. Looks like you've got it.

So, in going from $x=-1$ to $x=0$, for instance, the displacement is positive, even though $x$ is negative.

Ah I see, so one work is indeed positive (area 1) and the other one is negative (area 2) so the total is zero.

Thank you very much SammyS and hutchphd

 

[Orodruin]

Put simply: Displacement is the change in x, not x itself.