Ahockey puck is given an initial speed of 35 m/s on a frozen pond. the puck remains on the ice and slides 132 m before coming to rest. the acceleration of gravity is 9.8 m/s 2 . what is the coefficient of friction between the puck and the ice?

Using kinematic equation,
v²-u²=2as
v = final velocity = 0. since the puck comes to at rest
u = initial velocity = 35 m/s a = acceleration, s = distance traveled = 132 m
Plug all these values in above equation to get a,

0-35² = -2a(132)
35² = 2a(132)
a = 4.64 m/s² 

Net force acting on the puck is,
f = frictional force = μmg 
μ=coefficient of friction
According to Newton's 2nd law of motion,
f=ma
μmg = ma
μ=a/g = 4.64/9.8 = 0.473

B) as the temperature increases, the kinetic energy of the particles increases.|-/

Pe = 1/2k(x)^2  and .05n/m = .05x10^-9m = x