# Why work done by friction is positive? [closed]

Imagine a chain of mass m hanging from a table(not frictionless) which slips down the table. I want to calculate work done by friction. Obviously as it opposes slipping it should be negative. However while i solved it using integration my answer is positive. What am i missing? ## 2 Answers

You are mixing two things together.

The work computed by the integral $$W=\int \vec{F}_f \cdot d\vec{x}$$ is really an integral along some path. The dot product $$\vec{F}_f \cdot d\vec{x}$$ simply tells you "take the component of the force in the direction of the path and integrate along the path from initial point to final point as the distance rises". Now, if you take for simplicity just the endpoint of the chain, this endpoint moves from position $$x=l/4$$ to position $$x=0$$. The path integral bounds are however not from initial position to final position, but from initial distance traveled to end distance traveled. This makes sense, since in the case of constant force the work is indeed force times the distance traveled and not force times difference in position. Difference in position is of course negative, as the position is decreasing from $$l/4$$ to $$0$$, but the distance is positive and it grows from $$0$$ to $$l/4$$.

The problem is, you are calling both, the position and the distance along the path $$x$$, even though they go in opposite direction.

• @BioPhysicist what policy? Also, did you just downvoted for this sole reason? – Umaxo Aug 12 '20 at 13:03
• Check my work questions are off topic – BioPhysicist Aug 12 '20 at 13:09
• This answer helps though. Accepted. My doubt was addressed and resolved. How was it against the group policy? You downvoted my question too. @BioPysicist – Prakhar Pratap Mall Aug 12 '20 at 15:10
• @BioPhysicist It is not off topic and check my work type. It address my misconception of the concept of relating work by a force to position instead of distance. Please retract ur actions. – Prakhar Pratap Mall Aug 12 '20 at 15:25
• That is fine. You are entitled to have your opinion. Many other users disagree (which is why the question was closed). "help me spot my mistake" is exactly the same thing as "check my work". They are asking you to look at the work, and check what is wrong. When I see someone posting their work and asking others to point out what is wrong, to me that is clearly a check-my-work question, which is definitely off topic. – BioPhysicist Aug 13 '20 at 12:46

Limits taken are wrong. Since you have already taken cos180, you must take limits 0 to l/4, not the opposite. Then you'll get work done as negative

• But why. Limits are values of range of x. As x changes from l/4 to 0 the limits should also be that same l/4 to 0. – Prakhar Pratap Mall Aug 12 '20 at 5:59
• Initially, you have already assumed friction to be opposite to your direction of motion. If you had not assumed so, and taken cos0 instead, taking value of limits as l/4 to 0 would have been valid. You can think of the point at the edge of the table as the origin. – pyridine Aug 12 '20 at 6:04
• I do not understand "assumed friction to be in opposite direction". It is actuallly in opposite direction. How does assuming correct direction for the force invert the work done by it? Or are you suggesting that x ranges from -l/4 to 0 considering origin to be at the edge of the table? Does that mean any other choice of origin might give different results? – Prakhar Pratap Mall Aug 12 '20 at 6:18
• Please don't answer questions that blatantly go against site policy – BioPhysicist Aug 12 '20 at 12:55