if the coefficient of static friction between the tires and the road is 0.5. what is the shortest distance in which an automobile can be stopped while travellingat 15m/s
Answers
I don't know if I am absolutely right but please feel free to point out in case of a mistake.
So let the mass of the car be m, then the normal force,N acting upward is balanced by gravitational force, [math]F_g = mg[/math] acting downward.
Frictional force, [math]f=\myu*N [/math]where [math]\myu[/math] is the coefficient of friction.
Hence, [math]f = 0.5*m*g[/math]
Assuming the car is stopping due to frictional force only, we can write,
Deceleration, [math]a = \frac{f}{m} =\frac{-0.5*m*g}{m} = -0.5*g[/math]
Final velocity, v = 0
Initial velocity, u = 108kmph = 30m/s
From equation of motion, we have [math]v^2 - u^2 = 2*a*S[/math]
Taking acceleration due to gravity, [math]g = 9.8m /s^2[/math]
[math]S= \frac{-900}{2*(-0.5)*9.8} = \frac{900}{9.8} = 91.83m[/math]
Cheers!!!
Answer:
Lots of good answers here, but I prefer the easiest method to get the answer
work and energy methods
My first thought was that mass is not given, which means we can’t determine the normal force. And the normal force is required to calculate a friction force. This suggests to me that mass will cancel out in the equations which simply means the answer is independent of mass. I remember many of my students who didn’t know what to do without mass, so they arbitrarily set the mass = 1 kg. Still works out, I guess.
Getting back to my solution using energy methods. The initial kinetic energy is converted into heat due to friction to bring the object to a halt.
kinetic energy = work done by friction
KE=Ufric
or
KE=Ffricd
12mv2=(μFN)d
12mv2=(μmg)d
ta da ! mass cancels
12v2=μgd
Now just make sure all the units are consistent:
v=[ms]
d=[m]
g=[ms2]
v=(108000mhr)(hr3600sec)=30ms
12302=0.5(9.81)d
d=91.7 m
Technically the correct answer is d =90 m because it should only have 1 significant figure accuracy since the coefficient of friction only has 1 sig fig.