4. A hockey puck decelerates to a stop on the horizontal surface of an ice rink at a rate of -1.7 m/s. What is the coefficient of kinetic friction?

a. 0.173
b. 0.347
c. 5.76 m/s
d. 16.7

Puck friction to negative acceleration

fu=0.05

g = -9.81m/s^2

so

a = 0.05 * -9.81m/s^2 = -0.4905 m/s^2

For the moving puck

Velocity = v = u + (a*t) | u=initial velocity

So

Time = t = (v-u)/a

Distance = d = t *(v + u)/2

So

Distance d= ((v-u)/a) * ((v+u)/2)

This becomes d = (v-u)(v+u)/2a

Explanation:

Plugging in the numbers

d= (0 - 5.3)(0 + 5.3)/(2*-0.49.5) =28.63 meters

force tends to be low and long diustance masses are high