Question

In a shuffleboard game, the puck slides a total of 12 m before coming to rest. if the coefficient of kinetic friction between the puck and the horizontal board is 0.34, what was the initial speed of the puck?

Answer 1

The frictional force is given by

`F_f = mg \mu_D`
where m is the mass of the puck,
`g=9.81 m/s^2` and
`\mu_D=0.34` is the dynamic coefficient of friction.

The work done by this force to stop the puck is

`W=F_f d`
where
`d=12 m` is the total displacement of the puck.

The initial kinetic energy of the puck is

`K= (1)/(2) mv^2`
with v being the puck initial velocity. The puck comes to rest after 12 m, this means it loses all its kinetic energy, and for the principle of conservation of energy this loss of energy is equal to the work done by the frictional force. So we can write

`K=W`

`(1)/(2) mv^2 = mg\mu_D d`
and from this we can find the inizial velocity of the puck:

`v= √(2 g \mu_D d)= √(2(9.81 m/s^2)(0.34)(12 m))=8.95 m/s`