Braking force required to allow bike to move down slope with an arbitrary constant velocity 
Find the braking force required to allow a bike to move down a frictionless slope, inclined at $\theta$ radians above horizontal, with a constant velocity of v m/s.

I'm confused because Newton's 2nd Law of motion implies that the sum of forces on the bike must be zero in order for velocity to be constant, meaning that the braking force required is simply the bike's weight, resolved long the direction of the slope, in the opposite direction. But this only makes sure that velocity is constant.
How is it possible to apply a force to make the bike move with a constant velocity, of exactly v? My guess is that it has something to do with work, energy and power, but I haven't found a way yet.
 A: Draw a force diagram and determine the braking force required to counteract the (portion of) gravitational force pulling the bike down the slope.  If the net force is zero, delta-v will be zero.   I think the problem you were given assumes the bike starts out with velocity v .  If it doesn't then you'll need to derive a braking force whose magnitude varies as a function of velocity in some manner that goes asymptotic at the desired final velocity.
A: You are right that in this theoretical problem the breaking force to achieve constant velocity is the same for all velocities.  You either stipulate that the bike has the desired speed as a initial condition, or that it coasts with no breaking applied until the desired velocity is reached.
In the real world, there will always be some friction or drag forces that are dependent on velocity, so for a real bike going down a real hill, there will be different breaking forces required to keep the bike at different speeds.  Also there would be a human in the control loop adjusting the breaking pressure as needed to maintain the desired speed, probably without much conscious thought.  These factors together make the answer to this problem seem unintuitive, but you have it right.
