Question 5.12: The Sliding Hockey Puck A hockey puck on a frozen pond is gi...
The Sliding Hockey Puck
A hockey puck on a frozen pond is given an initial speed of 20.0 m/s. If the puck always remains on the ice and slides 115 m before coming to rest, determine the coefficient of kinetic friction between the puck and ice.
Conceptualize Imagine that the puck in Figure 5.20 slides to the right. The kinetic friction force acts to the left and slows the puck, which eventually comes to rest due to that force.
Categorize The forces acting on the puck are identified in Figure 5.20, but the text of the problem provides kinematic variables. Therefore, we categorize the problem in several ways. First, it involves modeling the puck as a particle under a net force in the horizontal direction: kinetic friction causes the puck to accelerate. There is no acceleration of the puck in the vertical direction, so we use the particle in equilibrium model for that direction. Furthermore, because we model the force of kinetic friction as independent of speed, the acceleration of the puck is constant. So, we can also categorize this problem by modeling the puck as a particle under constant acceleration.
Analyze First, let’s find the acceleration algebraically in terms of the coefficient of kinetic friction, using Newton’s second law. Once we know the acceleration of the puck and the distance it travels, the equations of kinematics can be used to find the numerical value of the coefficient of kinetic friction.
Apply the particle under a net force model in the x direction to the puck:
(1) \sum F_x=-f_k=m a_x
Apply the particle in equilibrium model in the y direction to the puck:
(2) \sum F_y=n-m g=0
Substitute n = mg from Equation (2) and f_k=\mu_k n into Equation (1):
\begin{aligned}-\mu_k n & =-\mu_k m g=m a_x \\a_x & =-\mu_k g\end{aligned}The negative sign means the acceleration is to the left in Figure 5.20. Because the velocity of the puck is to the right, the puck is slowing down. The acceleration is independent of the mass of the puck and is constant because we assume \mu_k remains constant.
Apply the particle under constant acceleration model to the puck, choosing Equation 2.17 from the model, v_{x f}{}^2=v_{x i}{ }^2+2 a_x\left(x_f-x_i\right), with x_i=0 and v_{x f}=0
v_{x f}{}^2=v_{x i}{ }^2+2 a_x\left(x_f-x_i\right) \quad \text { (for constant } a_x) (2.17)
0=v_{x i}{}^2+2 a_x x_f=v_{x i}{}^2-2 \mu_k g x_fSolve for the coefficient of kinetic friction:
\mu_k=\frac{v_{x i}{}^2}{2 g x_f}Substitute the numerical values:
\mu_k =\frac{(20.0 m/s)^2}{2\left(9.80 m/s^2\right)(115 m)}=0.177Finalize Notice that \mu_k is dimensionless, as it should be, and that it has a low value, consistent with an object sliding on ice.
