1. What is the coefficient of static friction between the box and the floor?
2. If the 48.0N force continues, the box accelerates at 0.70 m/s2. What is the coefficient of kinetic friction?
2006-09-28 19:34:07 · 4 answers · asked by bulldawg771 1 in Science & Mathematics ➔ Physics
This is no fun at all. It's a much better problem if it's an inclined plane instead of a flat floor ☺
1.
9.8*5*µs = 48N => µs = .976
2.
48N - 9.8*5*µk = 5*.7 => µk = .908
Doug
2006-09-28 19:50:18 · answer #1 · answered by doug_donaghue 7 · 0⤊ 0⤋
The limiting frictional force is 48 N
So mu * Mass * Acceleration due to gravity = 48 N
So mu * 5 * 9.8 = 48
mu = 48/49 = 0.979
Kinetic frictional force = (48 - 5*0.7) N = 44.5 N
Calculate the coefficient by yourself.
2006-09-28 19:44:02 · answer #2 · answered by astrokid 4 · 0⤊ 0⤋
1. u=(48N)/(5kg)(9.81m/s^2)
The maximum static friction force is the force required to get the object moving. That force is equal to the applied force. Divde that by the normal force wich is the mass and gravity.
2006-09-28 19:53:14 · answer #3 · answered by Anonymous · 0⤊ 0⤋
1. 0.98
2. 0.91
2006-09-28 20:01:08 · answer #4 · answered by entropy 3 · 0⤊ 0⤋