The problem involves calculating the work done by friction on a potted plant being moved along a shelf. The context includes a 2.75-kg mass and a coefficient of kinetic friction of 0.549, with a specified distance of 1.07 m.
The discussion is ongoing, with participants exploring different interpretations of the sign conventions related to work done by friction. Some guidance has been offered regarding the relationship between the direction of forces and displacement, but no consensus has been reached on the correct sign convention.
Participants are considering the implications of sign conventions in the context of work done, particularly regarding the direction of forces relative to displacement. There is an emphasis on understanding how these conventions affect the interpretation of the results.
rock.freak667 said:Well the work done by the frictional force would be positive due to sign conventions. So your answer would be W = + 15.8312 J, however the work done on the plant due to friction is W = -15.8312 J.
Your answer is correct, but the sign depends on the convention used.
You are asked for the work required to move the plant, i.e. the work done by the force that moves it. That force and the distance moved are in the same direction, so the product must be positive, no matter which way you assign the positive direction in the co-ordinate system. That is, either F and d are both negative or they're both positive, but Fd will always be positive.Lolagoeslala said:This would be negative correct?