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VARVARA [1.3K]
3 years ago
14

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
Physics
1 answer:
Komok [63]3 years ago
5 0

Answer:

The coefficient of static friction between the car and the track

u=0.572

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}=\frac{v_{t}^2}{r}

F_{fc}=m*\frac{v_{t}^2}{r}

But now vt relation with the tangential acceleration

v_{t}=2*a_{t}*\frac{\pi }{r}

replacing

F_{fc}=m*a_{t}*\frac{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*\sqrt{(1+\pi^2)}

u*m*g=m*a_{t}*(\sqrt{1+\pi^2})

Solve to u'

u=\frac{a_{t}}{g}*(\sqrt{1+\pi^2})

u=\frac{17\frac{m}{s^2} }{9.8\frac{m}{s^2}}*(\sqrt{1+\pi^2})=0.572

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Explanation:

Given that

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