A hockey puck is given an initial speed of 6.7 m/s. If the coefficient of kinetic friction between the puck and the ice is 0.08, how far (in m) does the puck slide before coming to rest?

Fk = u*Fn = 0.08*9.8M = 0.784M.

-0.784M = M*a.
a = -0.784 m/s^2.

V^2 = Vo^2 + 2a*d.
0 = 6.7^2 - 2*0.784d,
1.568d = 44.89,
d = 28.6 m.

To determine how far the puck slides before coming to rest, we need to calculate the distance covered by the puck using the equation of motion.

The key equation we can use for this problem is:

v^2 = u^2 + 2as

Where:
- v is the final velocity (0 m/s since the puck comes to rest)
- u is the initial velocity (6.7 m/s)
- a is the acceleration (given by the frictional force)
- s is the distance covered

The force of friction can be calculated using the equation:

F_friction = u * μ * m

where:
- F_friction is the force of friction
- u is the coefficient of kinetic friction (0.08)
- m is the mass of the puck (which is not given)

The acceleration can be calculated using Newton's second law:

F = m * a

Substituting the force of friction, we get:

F_friction = m * a
u * μ * m = m * a

Notice that the mass 'm' cancels out, so the mass of the puck is not needed to solve this problem.

Now, we can solve for acceleration:

a = u * μ

Substituting the values:

a = 6.7 m/s * 0.08
a = 0.536 m/s^2

Now, using the equation of motion, we can solve for the distance covered:

v^2 = u^2 + 2as
(0)^2 = (6.7)^2 + 2 * 0.536 * s

Rearranging the equation:

0 = 44.89 + 1.072s
-44.89 = 1.072s
s = -44.89 / 1.072
s ≈ -41.94 m/s

Since distance cannot be negative, we discard the negative value. The puck would slide approximately 41.94 meters before coming to rest.