Asked by mary
A still block slides over a still surface,if the mass of the pot is 2.00kg what is the force of friction opposing the motion?coefficient of kinetic friction is taken as 0.42.Gravity is 9.8
Answers
Answered by
Henry
Wb = mg = 2.0kg * 9.8N/kg = 19.6N. =
Weight of block.
Fb = 19.6N @ 0deg.
Fv = 19.6cos(0) = 19.6N. = Force perpendicular to the surface.
Fk = u*Fv = 0.42*19.6 = 8.23N. = Force of kinetic friction.
Weight of block.
Fb = 19.6N @ 0deg.
Fv = 19.6cos(0) = 19.6N. = Force perpendicular to the surface.
Fk = u*Fv = 0.42*19.6 = 8.23N. = Force of kinetic friction.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.