Force on a Cyclist: Net Force Calculation

[DarkEnergy890]

Homework Statement

A cyclist exerts a constant force of 450 N causing the bicycle to accelerate at 4.8 m s-2
until reaches a velocity of 20 m s-1 due west. Given that the combined mass of the
cyclist plus the bicycle is 120 kg, calculate (i) the net force (ii) the frictional force.

Relevant Equations

F=ma

My understanding is that the Net Force = Force Applied + Frictional Force. The net force is F=ma, so net force = 576N. Now 576N=450N + Frictional force, so frictional force has to be 126N.

My confusion is this:
1. Force is a vector, and the frictional force opposes the direction of motion. Therefore shouldn't friction turn out to be a negative?
2. If the applied force is 450N, then how can the net force be 576N, more than the force applied? If anything, I think that friction should cause the net force to be less than the applied force.

Here is the marking scheme (model answer) : https://prnt.sc/tyhQTEm0WKAw

Thanks!

 

[PeroK]

My confusion is this:
1. Force is a vector, and the frictional force opposes the direction of motion. Therefore shouldn't friction turn out to be a negative?

The model answer clearly show the friction acts east, which is a direction. Directions can be north, south, east, west etc. Not just $\pm$.

2. If the applied force is 450N, then how can the net force be 576N, more than the force applied? If anything, I think that friction should cause the net force to be less than the applied force.

A bicycle has pedals, gears and wheels that allow a force to be magnified, as it were. That's why it's possible to cycle uphill comfortably in low gear.

Here is the marking scheme (model answer) : https://prnt.sc/tyhQTEm0WKAw

That answer looks completely wrong to me. The force of the cyclist on the bicycle produces an equal and opposite force of the bicyle on the cyclist. These are therefore internal forces that cancel out.

The entire accelerating force must come from static friction of the tyres on the ground. Draw a free-body diagram and you'll see that the force from the ground on the bicycle must be $576 \ N$ (in the direction of the acceleration).

 

[berkeman]

That's why it's possible to cycle uphill comfortably in low gear.

Speak for yourself!

 

[haruspex]

As @PeroK observes, the question is poorly posed and the given solution is nonsense.
I'll assume this is on the horizontal (which should have been stated).

By "friction" the author appears to mean rolling resistance. Confusing that with friction is a novice error.

Saying that the rider exerts a force of 450N is pretty meaningless. The rider cannot be exerting a net horizontal force on the bike or the two would part company. The vertical force, mg, is irrelevant. If it means the force exerted on one pedal (no cleats, then) we would need to know about the gearing etc.
To make any use of it to solve the questions asked, we have to assume it means that, in the absence of losses, the cyclist would cause there to be a propulsive force of 450N on the bike+rider system. The actual force would come from static friction on the tyres acting forwards.

Finally, there is the sign error involving the "friction". Having written the equation as $F_{net}=F_{bike}-F_{friction}$, $F_{friction}$ must be being taken as positive to the East. Hence, getting -126N means the force is 126N West, aiding the acceleration!

Where does this sloppy problem come from?

 

[DarkEnergy890]

As @PeroK observes, the question is poorly posed and the given solution is nonsense.
I'll assume this is on the horizontal (which should have been stated).

By "friction" the author appears to mean rolling resistance. Confusing that with friction is a novice error.

Saying that the rider exerts a force of 450N is pretty meaningless. The rider cannot be exerting a net horizontal force on the bike or the two would part company. The vertical force, mg, is irrelevant. If it means the force exerted on one pedal (no cleats, then) we would need to know about the gearing etc.
To make any use of it to solve the questions asked, we have to assume it means that, in the absence of losses, the cyclist would cause there to be a propulsive force of 450N on the bike+rider system. The actual force would come from static friction on the tyres acting forwards.

Finally, there is the sign error involving the "friction". Having written the equation as $F_{net}=F_{bike}-F_{friction}$, $F_{friction}$ must be being taken as positive to the East. Hence, getting -126N means the force is 126N West, aiding the acceleration!

Where does this sloppy problem come from?

Thanks for the detailed reply! This problem came from a "pre-paper" company - not an official exam, so that's probably why there are so many mistakes in it.