Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 6 - Force and Motion-II - Problems: 42

Answer

13.4 m/s

Work Step by Step

When the car is on the verge of sliding, it means that the centripetal force is equal to the force of friction. The centripetal force is given by $F = ma_{c}$ where $a_{c} = v^2/r$ we know that r is the radius of the turn,which is 30.5m and we are looking for v. To find the Force of friction or Ff,max we know that it equals Ff = N*mu where mu is equal to 0.60. Then to find the Normal Force, since the only forces in the y direction are the weight of the car and the normal force and the car is not accelerating in the y direction then N = mg. Therefore, we have that $mv^2/(30.5) = 0.6 mg $ so that the masses cancel and we can solve for v, which we find to be 13.4 m/s
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