Question

Friction provides the force needed for a car to travel around a flat, circular race track. What is the maximum speed at which a car can safely travel if the radius of the track is 84.0 m and the coefficient of friction is 0.42?

Answer 1

`v=18.6 m/s`

Step-by-step explanation:

We need to equal the friction force and the centripetal force:

`F_(f)=F_(c)`

`\mu N=ma_(c)`

• N is the normal force (N = mg)
• μ is the coefficient of friction (0.42)
• a(c) is the centripetal force (a_{c}=v^{2}/R)

`\mu mg=m*(v^(2))/(R)`

`\mu g=(v^(2))/(R)`

`v=√(R\mu g)`

`v=√(84*0.42*9.81)`

`v=18.6 m/s`

I hope it helps you!

Answer 2

The maximum speed at which the car can safety travel around the track is 18.6m/s.

Step-by-step explanation:

Since the car is in circular motion, there has to be a centripetal force
`F_c`. In this case, the only force that applies for that is the static frictional force
`f_s`between the tires and the track. Then, we can write that:

`f_s=F_c`

And since
`f_s\leq \mu N` and
`F_c=(mv^(2))/(r)`, we have:

`\mu N\geq (mv^(2))/(r)`

Now, if we write the vertical equation of motion of the car (in which there are only the weight and the normal force), we obtain:

`N-mg=0\\\\\\implies N=mg`

Substituting this expression for
`N` and solving for
`v`, we get:

`\mu mg\geq (mv^(2))/(r)\\\\v\leq √(\mu gr)`

Finally, plugging in the given values for the coefficient of friction and the radius of the track, we have:

`v\leq \sqrt{(0.42)(9.81m/s^(2))(84.0m)}\\\\v\leq 18.6m/s`

It means that in its maximum value, the speed of the car is equal to 18.6m/s.