Question

A car traveling on a flat (unbanked) circular track accelerates uniformly from rest with a tangential acceleration of 1.7 m/s^2. The car makes it one-quarter of the way around the circle before it skids off the track. From these data, determine the coefficeint of static friction between the car and the track

Answer 1

The coefficient of static friction between the car and the track

u=0.572

Step-by-step explanation:

We don't know the mass of the car or any other information so the acceleration is the reason to solve the friction coefficient

∑
`F=F_(f)=m*a_(t)`

As we know

`F_(f)=u*F_(N)=u*m*g`

Also the center ward direction forces

`F_(fc)=m*a_(c)`

`a_(c)=(v_(t)^2)/(r)`

`F_(fc)=m*(v_(t)^2)/(r)`

But now vt relation with the tangential acceleration

`v_(t)=2*a_(t)*(\pi )/(r)`

replacing

`F_(fc)=m*a_(t)*(2\pi*r)/(2r)`

`F_(fc)=m*a_(t)*\pi`

So magnitude of the force can get by

`F_(f)=\sqrt{(m*a_(t)*\pi)^(2)+(m*a_(t))^(2)}`

Get the factor to simplify

`F_(f)=a_(t)*m*√((1+\pi^2))`

`u*m*g=m*a_(t)*(√(1+\pi^2))`

Solve to u'

`u=(a_(t))/(g)*(√(1+\pi^2))`

`u=(17(m)/(s^2) )/(9.8(m)/(s^2))*(√(1+\pi^2))=0.572`