Question

A hockey puck struck by a hockey stick is given an initial speed v0 in the positive x-direction. The coefficient of kinetic friction between the ice and the puck is µk.
(a) Obtain an expression for the acceleration of the puck. (Use the following as necessary: µk and g.)
a =
(b) Use the result of part (a) to obtain an expression for the distance d the puck slides. The answer should be in terms of the variables v0, µk, and g only.

Answer 1

(a) $a = -\mu_k g$

In order to find the acceleration of the puck, we must analyze the forces acting on it.

Along the vertical direction, we have two forces: the weight of the puck, mg, and the normal reaction, N. The two forces are equal and opposite so that the vertical acceleration is zero, so we can write

$N-mg = 0\rightarrow N = mg$ (1)

where m is the mass of the puck and g is the acceleration of gravity.

Along the horizontal direction, there is only one force acting on the puck: the force of friction, opposite to the direction of motion of the puck. So we can write:

$-\mu_k N = ma$ (2)

where $\mu_k$ is the coefficient of kinetic friction and a is the acceleration. Substituting (N) from eq.(1) into (2), we find an expression for a:

$-\mu_k mg = ma\\a = -\mu_k g$

(b) $d=\frac{v_0^2}{2\mu_kg}$

We can find the distance that the puck slides using the following SUVAT equation:

$v^2-v_0^2 = 2ad$

where

v = 0 is the final velocity of the puck (it comes to a stop, so v = 0)

$v_0$ is the initial velocity of the puck

a is the acceleration

d is the distance travelled

Substituting the expression for (a) that we found previously, we find an expression for d:

$d=\frac{-v_0^2}{2a}=\frac{-v_0^2}{-2\mu_k g}=\frac{v_0^2}{2\mu_kg}$