The free-body diagram shows four forces acting on the cyclist:
The net vertical force is the difference between the normal force (N) and the weight (W). Since the cyclist is not accelerating vertically, these forces must be equal. Therefore:
N = W = 650 N
The net horizontal force is the difference between the thrust (Th) and the drag force (D):
$F_{net,x} = Th - D = 120 N - 50 N = 70 N$
Since the net vertical force is zero, the magnitude of the net force is equal to the net horizontal force:
$|F_{net}| = \sqrt{F_{net,x}^2 + F_{net,y}^2} = \sqrt{70^2 + 0^2} = 70 N$
The net force is in the positive x-direction (to the right) since it's positive.
Since there is a net force acting on the cyclist, they are accelerating. A net force to the right means the cyclist is speeding up. We cannot calculate the _value_ of the speed without knowing the cyclist's mass and the time elapsed.
Magnitude of net force: \\(\boxed{70 \text{ N}}\\) Speed: increasing
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