(a) [tex]a = -\mu_k g[/tex]
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
[tex]N-mg = 0\rightarrow N = mg[/tex] (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:
[tex]-\mu_k N = ma[/tex] (2)
where [tex]\mu_k[/tex] is the coefficient of kinetic friction and a is the acceleration. Substituting (N) from eq.(1) into (2), we find an expression for a:
[tex]-\mu_k mg = ma\\a = -\mu_k g[/tex]
(b) [tex]d=\frac{v_0^2}{2\mu_kg}[/tex]
We can find the distance that the puck slides using the following SUVAT equation:
[tex]v^2-v_0^2 = 2ad[/tex]
where
v = 0 is the final velocity of the puck (it comes to a stop, so v = 0)
[tex]v_0[/tex] 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:
[tex]d=\frac{-v_0^2}{2a}=\frac{-v_0^2}{-2\mu_k g}=\frac{v_0^2}{2\mu_kg}[/tex]
