Coefficient of kinetic friction on car

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Homework Help Overview

The discussion revolves around a physics problem involving a 500 kg car being towed on a flat surface, where participants are tasked with finding the friction force and the coefficient of kinetic friction given the acceleration and the towing force.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore Newton's second law to relate forces and acceleration, with some attempting to derive the frictional force from the given data. Questions arise regarding the calculations and the interpretation of the forces involved.

Discussion Status

Some participants have provided insights into the relationship between net force, mass, and acceleration, while others express confusion about specific calculations and the reasoning behind them. There is an ongoing exploration of the problem without a clear consensus on the approach to take.

Contextual Notes

Participants are navigating through the problem with varying levels of understanding, and there are indications of assumptions regarding the forces acting on the car and the definitions of friction coefficients. The discussion highlights the challenges of interpreting the problem setup and calculations.

unknownfrost
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Ok here is my situation. There is a problem that states:

Find the friction force if the acceleration of a 500 kg car being towed on a flat surface is 5 m/s2 and a tow truck is pulling with a force of 3550 N.

I do not where to go from here. Furthermore, the next question asks:

What would the coefficient of kinetic friction be?

If someone could help me out with this I would be extremely thankful!

This has got me going :eek: and :cry:
 
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First, notice that, according to Newtons 2nd law, if the net force on the truck were to be 3550 N, the acceleration of the truck would be:
[tex]a=\frac{F}{m}=\frac{3550}{500}=7.1 m/s^2[/tex]
So the frictional force is of that magnitude to bring the acceleration down to [itex]5 m/s^2[/itex].
Can you find the force needed to do this?
(There's a slightly quicker way. But I find this more instructive)
 
Well...
I know that f = (mk)(n)
Would I use the 3500 from earlier and take mk = f/n to get the coeff. of kinetic?
 
its a simple force diagram

ma = (Force tow truck) - Friction

(500kg)(5m/s^2) = 3500 N - uN
2500 N - 3500 N = -1000 = -umg

1000 N/((500kg)(9.81 m/s^2)) = u = 0.220 < friction coffecient
 
Phymath said:
its a simple force diagram

ma = (Force tow truck) - Friction

(500kg)(5m/s^2) = 3500 N - uN
2500 N - 3500 N = -1000 = -umg

1000 N/((500kg)(9.81 m/s^2)) = u = 0.220 < friction coffecient



ok I'm with you on the 3500 part, but where did you get the 2500 from?

I understand now how to plug it in. i just don't know how you got it!

I could just take the awnser but I am totally lost on the awnser!
Physics is tough man!
 
Last edited:
Think about it like this.
You are given that a 500 kg object has an acceleration of 5 m/s^2.
Then Newton's law says: "Aha, it accelerates. So there must be a net force acting on it."
This net force is equal to its mass times its acceleration: [itex]F=ma=500 \cdot 5 = 2500 N[/itex].
So the net force on the car is 2500 N. Since the tow truck is pulling with a force of 3550 N, something must be exerting a force in the other direction (which is the force of friction). Can you take it from here?
 

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