Question

A hockey puck is hit on a frozen lake and starts moving with a speed of 12.0 m/s. exactly 5.0 s later, its speed is 6.0 m/s. what is the puck's average acceleration? what is the coefficient of kinetic friction between the puck and the ice?

Answer 1

1) In an uniformly accelerated motion, the acceleration is given by:
$a= \frac{v_f-v_i}{\Delta t}$
where $v_f$ is the final speed, $v_i$ is the initial speed, and $\Delta t$ is the time interval between the initial and final point of the motion.

Using the data of the problem: $v_i = 12.0 m/s$, $v_f = 6.0 m/s$, and the time $\Delta t = 5.0 s$, the acceleration is
$a= \frac{6 m/s-12 m/s}{5.0 s}=-1.2 m/s^2$
where the negative sign means that the hockey puck is decelerating.

2) The frictional force F between the puck and ice is responsible for the deceleration of the puck, and for second Newton's law this force is equal to the product between the mass of the puck m and the acceleration:
$F=ma$ (1)
The frictional force can also be written as
$F=-\mu_D mg$ (2)
where $\mu_D$ is the coefficient of kinetic friction, and where the negative sign is due to the fact that the frictional force acts against the direction of motion of the puck. By equilizing (1) and (2) we can find the value of this coefficient:
$ma=-\mu_D mg$
$\mu_D = -\frac{a}{g}= -\frac{-1.2 m/s^2}{9.81 m/s^2}=0.12$