Asked by ellimac
a 1.4 kg block slides across a rough surface such that it slows down with an acceleration of 1.25 m/s2. What is the coefficient of kinetic frisction between the block and the surface?
Answers
Answered by
drwls
I assume that the surface is horizontal. The deceleration rate is -F/M. So, the friction force is
a = -1.4/1.25 = -1.2 Newtons
The coefficient of sliding friction is muk = F/(M*g) = 1.2/(1.4*9.8) = ?
a = -1.4/1.25 = -1.2 Newtons
The coefficient of sliding friction is muk = F/(M*g) = 1.2/(1.4*9.8) = ?
Answered by
Aths'er
Given: m= 1.4 kg
A= -1.25 m/s2
µk=Kinetic force/ Normal Force
Normal Force= 1.4*9.81(gravity) = 13.72 N
Kinetic force = net force
net force = ma = 1.4(-1.25) = -1.75
µk= -1.75/13.72 = -0.13
A= -1.25 m/s2
µk=Kinetic force/ Normal Force
Normal Force= 1.4*9.81(gravity) = 13.72 N
Kinetic force = net force
net force = ma = 1.4(-1.25) = -1.75
µk= -1.75/13.72 = -0.13
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.