A smooth 0.165kg hockey puck slides along a smooth floor at an initial speed of 1.0m/s and stops in a distance of 2.26m find the work done by the normal force that the floor exerts upward on the puck. Calculate the work done by friction.

So I found that the work done should be zero since the force of gravity and the force of normal (which is the only force acting on it) is perpendicular to the displacement of the puck (since the puck is sliding on a smooth floor).

What I'm having trouble with is finding the work done by friction. This is a smooth floor which should mean there is no friction at all. If anybody can give a nice push into the correct direction I would be forever grateful.

The floor may be smooth, but it's not frictionless - if it was frictionless the puck would slide forever without slowing, and that's not what happens in this problem.

Your answer to the first part is correct, so you understand the concept of work being force times distance traveled in the direction of that force. Here friction opposes the motion of the puck, and so is directly in line with that motion. To get started you're going to have to calculate the acceleration of the puck from the data provided, and from that find the force that causes that acceleration (F=ma).

Or another approach - consider that the initial kinetic energy of the puck must equal the work required to stop it. I suggest that you try both approaches and see how they work out - it will be good practice for you.