Asked by Kim
A force of 36.1 N is required to start a 4.0 -kg box moving across a horizontal concrete floor. If the 36.1 N force continues, the box accelerates at 0.71 m/s^2. What is the coefficient of kinetic friction?
I got 0.072 as my answer and this is showing as incorrect.
I did:
36.1 - mu x 4.0kg x 9.8 = 4.0kg x 0.71
0.71 / 9.8 = 0.072
I got 0.072 as my answer and this is showing as incorrect.
I did:
36.1 - mu x 4.0kg x 9.8 = 4.0kg x 0.71
0.71 / 9.8 = 0.072
Answers
Answered by
bobpursley
YOur math is amazing. Next time, follow rules of algebra.
36.1-mu*4*9.8=4*.71
mu=(36.1-4*.71)/(4*9.8)=.85 check that.
36.1-mu*4*9.8=4*.71
mu=(36.1-4*.71)/(4*9.8)=.85 check that.
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.