Coefficient of kinetic Friction given velocity and distance.

AI Thread Summary
To find the coefficient of kinetic friction for a car sliding to a stop, the key is to apply Newton's second law and the friction force equation. The car's initial velocity is 20 m/s, and it comes to a stop over a distance of 40 m. The normal force can be expressed as N = mg, allowing the mass to be treated as a variable that cancels out in the calculations. The frictional force, which is responsible for deceleration, can be calculated using the car's deceleration derived from its initial velocity and stopping distance. Ultimately, the coefficient of kinetic friction can be determined without knowing the mass, leading to the conclusion that it is approximately 0.51.
pech0706
Messages
13
Reaction score
0

Homework Statement


A car is moving with a velocity of 20m/s when the brakes are applied and the wheels lock (stop spinning), The car then slides to a stop in 40m. FInd the coefficient of kinitic friction between the tires and the road.



Homework Equations


F=ma
Ffriction=coefficient fo friction*N
N=mg

The Attempt at a Solution


I have drawn a picture and free body diagram, but that's as far as I've gotten. I'm stuck because I don't know how to calculate the normal force without the mass, and I don't now how to find the coefficient without N. I know the answer is 0.51, but I do not know how to get there.
 
Physics news on Phys.org
dont worry about mass.
take it as m
it will get canceled in the process.
 
So, basically that means the total force acting on the car is the velocity and the frictional force. After ignoring the mass though, I'm still stuck. I don't know how to find the frictional force or the coefficent of friciton when ignoring mass.
 
Dont ignore mass, take it as m.

Then do the general method and find the answer.
mass won't be any problem
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top