Question
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
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.