Answer:
the 40 m hill
Explanation:
because the longer it goes downhill the more speed it will pick up.
The net force on the block acting perpendicular to the incline is
∑ <em>F</em> = <em>n</em> - <em>w</em> cos(29.4°) = 0
where <em>n</em> is the magnitude of the normal force and <em>w</em> = <em>m g</em> is the weight of the block.
The equation itself comes from splitting up the forces acting on the block into components pointing parallel or perpendicular to the incline. The only forces acting on the block in the perpendicular direction are the normal force and the perpendicular component of the block's weight.
Solve for <em>n</em> :
<em>n</em> = <em>m g</em> cos(29.4°)
<em>n</em> = (6 kg) (9.80 m/s²) cos(29.4°)
<em>n</em> ≈ 51.2 N
I think that the angular velocity vector points at right angles to the direction in which the wheels are turning (spindle on an old fashioned record player ?) and so at right angles to the direction the bike is moving in. This contributes to the gyroscope effect on the wheels and bike and allows a rapidly rotating wheel to be more stable than a slowly rotating one. Problem for the trainee cyclist is to believe that they are actually more stable when their wheels are moving quickly. 'cos the tendency is to go slowly to start with, which makes balancing harder.
But then, most cyclists, especially youngsters, don't sit down all day analysing circular motion vectors, which may be just as well.