Answer:
The coefficient of kinetic friction between the puck and the ice is 0.11
Explanation:
Given;
initial speed, u = 9.3 m/s
sliding distance, S = 42 m
From equation of motion we determine the acceleration;
v² = u² + 2as
0 = (9.3)² + (2x42)a
- 84a = 86.49
a = -86.49/84
|a| = 1.0296
= ma
where;
Fk is the frictional force
μk is the coefficient of kinetic friction
N is the normal reaction = mg
μkmg = ma
μkg = a
μk = a/g
where;
g is the gravitational constant = 9.8 m/s²
μk = a/g
μk = 1.0296/9.8
μk = 0.11
Therefore, the coefficient of kinetic friction between the puck and the ice is 0.11
We know that a=vf_vi/t equals equation "a" . Where a is the acceleration of the body , vf is the final velocity , vi is the initial velocity and t is equal to time . Since vi equals o m/s , vf equals to 60 m/s and t equals 10 s. Put in equation "a". a=60-0/10 =6m/s2
Newton has 3 Laws specifically The Three Laws of Motion
60,000 is 100 times as much as 600