Solve Friktion & Effect Question: Alex on Motorcycle

[anthroxy]

Question

Alex i sitting on a motorcycle, the wheel goes makes 95 turns per minute. The distance between center of the wheel to the wheels end is 0,25m (which gives us the radius) but is affected by a breaking force of 28N, how big is the work of friktion during the 45min? and what is the effect during this time?

My attempt to a solution

Since we have the radius of the wheel we can find out the wheels perimeter combined with the amount turns per minute and the time.

Perimiter of a cirlcle is = π2r
so, π2(0,25) = 1,57m now it does 1,57m ninetyfive times per minute thus 95 * 1,57 = 149,15m which is the distance covered per minute so 149,15 * 45 = 6711,75m which is the distance covered over 45min

S = 6711,75m

since we now have the distance we can get the velocity using v = d/t (and i converted 45min into seconds) therefore v = 6711,75/2700 hence we get v = 2,48m/s

But this is where i get stuck... =/

 

[JaredJames]

I'm assuming this question requires you to calculate the torque on the wheel from the friction.

Torque = Force x Radius

Aside from that, I see nothing else to add.

If the friction is greater than driving force it will slow down.
If the friction is equal to the driving force it will remain steady.
If the friction is less than the driving force it will accelerate.

Otherwise I'm not sure. I don't understand the wording of the question.

 

[CxStrike]

What equation are you using that requires you to get the velocity? If it's a work problem, then W=Fd. You're given force, and you've solved for distance. Now just multiply to get work.

 

[anthroxy]

What equation are you using that requires you to get the velocity? If it's a work problem, then W=Fd. You're given force, and you've solved for distance. Now just multiply to get work.

But there is a breaking force of 28N and isn't there a friction force on top of that?

i thought that once i had the friction force i would add it together with the breaking force which would be the total amount of force slowing down the wheel and then multiply that force with the distance traveled to get the work for total 45min and from there i could just simply divide the work by time to get the total effect during the 45min.

But my problem lies with the friction force i don't know how to calculate it, does my explanation make sense?

 

[CxStrike]

But there is a breaking force of 28N and isn't there a friction force on top of that?

i thought that once i had the friction force i would add it together with the breaking force which would be the total amount of force slowing down the wheel and then multiply that force with the distance traveled to get the work for total 45min and from there i could just simply divide the work by time to get the total effect during the 45min.

But my problem lies with the friction force i don't know how to calculate it, does my explanation make sense?

The friction force is the braking force, I believe. Unless you are talking about the friction between the wheel and the ground, but to find that you'd need the weight.