asteringPhysics: HWs Ch. 5 Google Chrome Secure https//session.masteringphysics.
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asteringPhysics: HWs Ch. 5 Google Chrome Secure https//session.masteringphysics.com/myct/itemView?assignmentProblemID-83594275 HW5- Ch.5 Exercise 5.48 « previous | 7 of 13 |next Part A Exercise 5.48 What is the minimum ooefficient of static friction that will prevent sliding? A flat (unbanked) curve on a highway has a radius of 250.0 m. A car rounds the curve at a speed of 27.0 m/ umin = 1.068 Submit My Answers ive Up Incorrect; Try Again; 4 attempts remaining Part B Suppose that the highway is icy and the coefficient of friction between the tires and pavement is only one-third what you found in the previous part. What should be the maximum speed of the car so it can round the curve safely? Express your answer with the appropriate units Value Units Submit My Answers Give Up Continue 9:34 PM lype here to search 2 10/6/2017 6Explanation / Answer
1)
Cars turning around curves, can only do so if they don't exceed fsMax (static friction Max)
fsMax = N ( N is normal force and is coefficient of static friction)
You know that there is a radial force, Fr= fs = (mv2)/r
and a Vertical force (Normal force) to cancel out the gravity of course, Fz= N-mg = 0
the normal force and gravity have to cancel since it's not moving up and down.
the reason why the radial force is equal to fs is that friction is the only thing that is letting it maintain your control on the slick road.
since fs = N
N = (mv2)/r
and since it is unbanked mg = N
So that, we get ==> mg = (mv2)/r ------------------(eq 1)
= v2/(gr) =(27.0m/s/{(9.81m/s²)(250m)} = 0.297
2) from eq 1 ==> Vmax = (rg)
Here is one-third of previous part so = (0.297)/3 = 0.09
==> Vmax = ([(0.099)(9.81m/s²)(250m)] = 15.58m/s
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