These are all basically on the same concept so I'll help you out with this one and ask you to try the rest again. If you still need help, please let us know in this thread.

The weird thing about this problem is the constant frictional force. Typically we would have a constant coefficient of friction. Strange.

The picture I have is that the slope is angling downward to the right. I am choosing a coordinate system for this part to have +x down the slope and +y perpendicular (and upward) to the slope. The Free Body Diagram shows three forces on the cyclist: A normal force (n) in the +y direction, a weight (w) acting straight down, and a kinetic friction force (f) acting up the slope.

On the horizontal surface we have a similar situation. I am taking +x to the right and +y upward. The FBD has a normal force (N) acting in the +y direction, a friction force (f) the same magnitude as before in the -x direction, and a weight (w) acting straight downward. (Note that the normal force here is different than the normal force on the slope.)

It will be simpler here to use the work-energy theorem rather than Newton's 2nd Law. (Typically we would use the latter approach.)

Picking the starting point to be at the top of the slope and the ending point to be where the cyclist stops is probably the easiest way since we don't care what happens in between. I am choosing the 0 point for the gravitational potential energy to be at the level where the cyclist is on the horizontal part of the motion.

The friction is constant for both parts so the nonconservative work done by it is simply

Thus

You can do the rest from here.

-Dan